Question
1.calculate the hydrogen concentration for each of the following.
(a).blood plasma, pH=7.4
(b).orange juice, pH=3.5
(c).human urine, pH=6.2
(c).gastric juice, pH=1.8
(d).household ammonia, pH=11.5
2. what buffer is used in the body system to preserve the following
a)urine
b)saliva
c)egg white
(a).blood plasma, pH=7.4
(b).orange juice, pH=3.5
(c).human urine, pH=6.2
(c).gastric juice, pH=1.8
(d).household ammonia, pH=11.5
2. what buffer is used in the body system to preserve the following
a)urine
b)saliva
c)egg white
Answers
Answer
2. What is the [πΆπ»_3 πΆππ^β ]/[πΆπ»_3 πΆπππ»] ratio in an acetate buffer at pH 5.00
Answer
3. How would the following buffers differ in their capacity
a).0.01M γππγ_2 π»ππ_4 πππ 0.01π πππ»_2 ππ_4
b).0.1M γππγ_2 π»ππ_4 πππ 0.1π πππ»_2 ππ_4
c).1.0M γππγ_2 π»ππ_4 πππ 1.0π πππ»_2 ππ_4
a).0.01M γππγ_2 π»ππ_4 πππ 0.01π πππ»_2 ππ_4
b).0.1M γππγ_2 π»ππ_4 πππ 0.1π πππ»_2 ππ_4
c).1.0M γππγ_2 π»ππ_4 πππ 1.0π πππ»_2 ππ_4
Answer
how could you know that this is buffer pairs if you have been given a multiple choice question test? give some examples
Answer
4. Calculate the [π»^+ ] πππ [ππ_3^( β) ] of a 0.175M solution of nitric acid, ππ»π_3^( β).
5. Calculate the hydroxide ion concentration in a 0.175 solution of HCl.
6. Calculate the hydrogen ion concentration in 0.01M solution of Ca(ππ»)_2
7. The ionisation constant of water is πΎ_π€=1.00Γ10^(β13.60) at body temperature, 37^0 πΆ. πβππ‘ πππ π‘βπ π»_3 π^+ ππγ ππ»γ^β concentrations at that temperature?
8. Calculate the pH of the following solutions at 25^0 πΆ:(π)0.0028π π»πΆπ, (π)0.014 π»ππ_3, (π)0.00052π π»π΅π,
0.092π π»πΌ
9. Calculate the pH of a 0.025M solution of ammonia, πΎ_π=1.75Γ 10^(β5)
5. Calculate the hydroxide ion concentration in a 0.175 solution of HCl.
6. Calculate the hydrogen ion concentration in 0.01M solution of Ca(ππ»)_2
7. The ionisation constant of water is πΎ_π€=1.00Γ10^(β13.60) at body temperature, 37^0 πΆ. πβππ‘ πππ π‘βπ π»_3 π^+ ππγ ππ»γ^β concentrations at that temperature?
8. Calculate the pH of the following solutions at 25^0 πΆ:(π)0.0028π π»πΆπ, (π)0.014 π»ππ_3, (π)0.00052π π»π΅π,
0.092π π»πΌ
9. Calculate the pH of a 0.025M solution of ammonia, πΎ_π=1.75Γ 10^(β5)
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1. To calculate the hydrogen concentration (H+) for each solution, we can use the formula:
[H+] = 10^(-pH)
(a) Blood plasma, pH = 7.4
[H+] = 10^(-7.4)
[H+] β 3.98 x 10^(-8) M
(b) Orange juice, pH = 3.5
[H+] = 10^(-3.5)
[H+] β 3.16 x 10^(-4) M
(c) Human urine, pH = 6.2
[H+] = 10^(-6.2)
[H+] β 1.58 x 10^(-7) M
(d) Gastric juice, pH = 1.8
[H+] = 10^(-1.8)
[H+] β 1.58 x 10^(-2) M
(e) Household ammonia, pH = 11.5
[H+] = 10^(-11.5)
[H+] β 3.16 x 10^(-12) M
2. The buffers used in the body system to preserve different substances are:
(a) Urine: The buffer system used in urine is the phosphate buffer system, which consists of the dihydrogen phosphate ion (H2PO4-) and hydrogen phosphate ion (HPO4^2-).
(b) Saliva: The buffer system used in saliva is the bicarbonate buffer system, which consists of the bicarbonate ion (HCO3-) and carbonic acid (H2CO3).
(c) Egg white: The buffer system used in egg white is the albumin buffer system, where the protein albumin acts as a buffer to maintain the pH stability.
[H+] = 10^(-pH)
(a) Blood plasma, pH = 7.4
[H+] = 10^(-7.4)
[H+] β 3.98 x 10^(-8) M
(b) Orange juice, pH = 3.5
[H+] = 10^(-3.5)
[H+] β 3.16 x 10^(-4) M
(c) Human urine, pH = 6.2
[H+] = 10^(-6.2)
[H+] β 1.58 x 10^(-7) M
(d) Gastric juice, pH = 1.8
[H+] = 10^(-1.8)
[H+] β 1.58 x 10^(-2) M
(e) Household ammonia, pH = 11.5
[H+] = 10^(-11.5)
[H+] β 3.16 x 10^(-12) M
2. The buffers used in the body system to preserve different substances are:
(a) Urine: The buffer system used in urine is the phosphate buffer system, which consists of the dihydrogen phosphate ion (H2PO4-) and hydrogen phosphate ion (HPO4^2-).
(b) Saliva: The buffer system used in saliva is the bicarbonate buffer system, which consists of the bicarbonate ion (HCO3-) and carbonic acid (H2CO3).
(c) Egg white: The buffer system used in egg white is the albumin buffer system, where the protein albumin acts as a buffer to maintain the pH stability.
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The [πΆπ»_3 πΆππ^β ]/[πΆπ»_3 πΆπππ»] ratio in an acetate buffer can be determined using the Henderson-Hasselbalch equation:
ππ» = ππΎπ + πππ ([πΆπ»_3 πΆππ^β ]/[πΆπ»_3 πΆπππ»])
Given that pH = 5.00 and the pKa value for acetic acid is 4.76 (approximately), we can rearrange the equation to:
[πΆπ»_3 πΆππ^β ]/[πΆπ»_3 πΆπππ»] = 10^(ππ» β ππΎπ)
[πΆπ»_3 πΆππ^β ]/[πΆπ»_3 πΆπππ»] = 10^(5.00 - 4.76)
[πΆπ»_3 πΆππ^β ]/[πΆπ»_3 πΆπππ»] β 1.78
Therefore, the [πΆπ»_3 πΆππ^β ]/[πΆπ»_3 πΆπππ»] ratio in the acetate buffer at pH 5.00 is approximately 1.78.
ππ» = ππΎπ + πππ ([πΆπ»_3 πΆππ^β ]/[πΆπ»_3 πΆπππ»])
Given that pH = 5.00 and the pKa value for acetic acid is 4.76 (approximately), we can rearrange the equation to:
[πΆπ»_3 πΆππ^β ]/[πΆπ»_3 πΆπππ»] = 10^(ππ» β ππΎπ)
[πΆπ»_3 πΆππ^β ]/[πΆπ»_3 πΆπππ»] = 10^(5.00 - 4.76)
[πΆπ»_3 πΆππ^β ]/[πΆπ»_3 πΆπππ»] β 1.78
Therefore, the [πΆπ»_3 πΆππ^β ]/[πΆπ»_3 πΆπππ»] ratio in the acetate buffer at pH 5.00 is approximately 1.78.
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The buffer capacity refers to the ability of a buffer solution to resist changes in pH when small amounts of acid or base are added. It is directly influenced by the concentration of the buffer components.
(a) 0.01M Na2HPO4 and 0.01M NaH2PO4 buffer:
The buffer capacity is relatively low in this case because the concentration of the buffer components is relatively low.
(b) 0.1M Na2HPO4 and 0.1M NaH2PO4 buffer:
The buffer capacity is higher in this case compared to (a) because the concentration of the buffer components is higher. Therefore, it can resist pH changes caused by the addition of small amounts of acid or base better.
(c) 1.0M Na2HPO4 and 1.0M NaH2PO4 buffer:
The buffer capacity is the highest in this case because the concentration of the buffer components is the highest. It can effectively prevent significant pH changes even with greater amounts of acid or base added.
(a) 0.01M Na2HPO4 and 0.01M NaH2PO4 buffer:
The buffer capacity is relatively low in this case because the concentration of the buffer components is relatively low.
(b) 0.1M Na2HPO4 and 0.1M NaH2PO4 buffer:
The buffer capacity is higher in this case compared to (a) because the concentration of the buffer components is higher. Therefore, it can resist pH changes caused by the addition of small amounts of acid or base better.
(c) 1.0M Na2HPO4 and 1.0M NaH2PO4 buffer:
The buffer capacity is the highest in this case because the concentration of the buffer components is the highest. It can effectively prevent significant pH changes even with greater amounts of acid or base added.
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In a multiple-choice question test, there are several indications that a pair of substances could be a buffer system. Some of these indications include:
1. The presence of both a weak acid and its conjugate base or a weak base and its conjugate acid in the given options. Buffer systems commonly consist of a weak acid/base and its conjugate pair.
Example:
Option A: HCl and NaCl (not a buffer pair)
Option B: H2CO3 and NaHCO3 (buffer pair)
Option C: HNO3 and NaNO3 (not a buffer pair)
Option D: HCN and NaCN (buffer pair)
In this example, options B and D contain a weak acid (H2CO3 and HCN) and their corresponding conjugate base (NaHCO3 and NaCN), making them potential buffer pairs.
2. The option includes a salt of a weak acid and a strong base or a salt of a weak base and a strong acid. Salt solutions can act as buffer systems to some extent.
Example:
Option A: NaOH and HCl (not a buffer pair)
Option B: Na2CO3 and HCl (buffer pair)
Option C: NH3 and HNO3 (not a buffer pair)
Option D: NH4Cl and NaOH (buffer pair)
In this example, options B and D contain a salt of a weak acid (Na2CO3) and a strong acid (HCl) or a salt of a weak base (NH4Cl) and a strong base (NaOH), making them potential buffer pairs.
It's important to keep in mind that other factors, such as pH values and concentrations, may also need to be considered to determine the presence of a buffer system accurately.
1. The presence of both a weak acid and its conjugate base or a weak base and its conjugate acid in the given options. Buffer systems commonly consist of a weak acid/base and its conjugate pair.
Example:
Option A: HCl and NaCl (not a buffer pair)
Option B: H2CO3 and NaHCO3 (buffer pair)
Option C: HNO3 and NaNO3 (not a buffer pair)
Option D: HCN and NaCN (buffer pair)
In this example, options B and D contain a weak acid (H2CO3 and HCN) and their corresponding conjugate base (NaHCO3 and NaCN), making them potential buffer pairs.
2. The option includes a salt of a weak acid and a strong base or a salt of a weak base and a strong acid. Salt solutions can act as buffer systems to some extent.
Example:
Option A: NaOH and HCl (not a buffer pair)
Option B: Na2CO3 and HCl (buffer pair)
Option C: NH3 and HNO3 (not a buffer pair)
Option D: NH4Cl and NaOH (buffer pair)
In this example, options B and D contain a salt of a weak acid (Na2CO3) and a strong acid (HCl) or a salt of a weak base (NH4Cl) and a strong base (NaOH), making them potential buffer pairs.
It's important to keep in mind that other factors, such as pH values and concentrations, may also need to be considered to determine the presence of a buffer system accurately.
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4. To calculate the [π»^+ ] and [ππ_3^(-) ] of a 0.175M solution of nitric acid, ππ»π_3^(-), we need to consider the dissociation of nitric acid in water:
ππ»π_3 -> π»^+ + ππ_3^(-)
Since nitric acid is a strong acid, it will completely dissociate in water. Therefore, the initial concentration of π»^+ ions will be equal to the initial concentration of nitric acid, which is 0.175M. The concentration of ππ_3^(-) ions will be zero since it is a spectator ion and does not participate in the formation of additional ππ»π_3 molecules.
[π»^+ ] = 0.175M
[ππ_3^(-) ] = 0M
5. In a 0.175M solution of HCl, HCl is a strong acid that will dissociate completely in water:
π»πΆπ -> π»^+ + πΆπ^(-)
Since HCl is a strong acid, the initial concentration of π»^+ ions will be equal to the initial concentration of HCl, which is 0.175M. The concentration of hydroxide ions (OH-) in this acidic solution will be negligible.
[π»^+ ] = 0.175M
[ππ»^- ] β 0M
6. In a 0.01M solution of Ca(OH)2, Ca(OH)2 is a strong base that will dissociate completely in water:
Ca(OH)2 -> 2ππ»^- + πΆπ^2+
Each Ca(OH)2 molecule dissociates to produce two hydroxide ions (OH-), and the concentration of hydroxide ions will be twice the initial concentration of Ca(OH)2.
[π»^+ ] β 0M
[ππ»^- ] = 2 * 0.01M = 0.02M
7. The ionization constant of water, πΎ_π€, can be expressed as:
πΎ_π€ = [π»_3π^+ ]*[ππ»^- ]
At body temperature, 37Β°C, the value of πΎ_π€ is 1.00 x 10^(-13.60). To determine the concentrations of [π»_3π^+ ] and [ππ»^- ], you would need to calculate the square root of πΎ_π€ since the concentration of [π»_3π^+ ] and [ππ»^- ] will be equal:
[π»_3π^+ ] = [ππ»^- ] = πΎ_π€^(0.5) = (1.00 x 10^(-13.60))^(0.5)
Simply calculate the square root of πΎ_π€ to determine the concentrations at body temperature.
8. To calculate the pH of the following solutions at 25Β°C:
(a) 0.0028M HCl:
Since HCl is a strong acid that dissociates completely, the concentration of H+ ions will be equal to the initial concentration of HCl. Use the formula for pH:
pH = -log[H+]
pH = -log(0.0028)
(b) 0.014M HNO3:
Since HNO3 is also a strong acid that dissociates completely, the concentration of H+ ions will be equal to the initial concentration of HNO3. Use the formula for pH:
pH = -log[H+]
pH = -log(0.014)
(c) 0.00052M HBr:
HBr is a strong acid, so the concentration of H+ ions will be equal to the initial concentration of HBr. Use the formula for pH:
pH = -log[H+]
pH = -log(0.00052)
(d) 0.092M HI:
HI is a strong acid, so the concentration of H+ ions will be equal to the initial concentration of HI. Use the formula for pH:
pH = -log[H+]
pH = -log(0.092)
9. To calculate the pH of a 0.025M solution of ammonia, use the given Kb value:
Kb = [OH-][NH4+]/[NH3]
Since ammonia (NH3) is a weak base, we can assume that the concentration of [OH-] is negligible compared to [NH3]. Therefore, we can simplify the equation to:
Kb β [OH-] * [NH4+]/[NH3]
To determine [OH-] concentration, you can use the equation:
[OH-] = Kb * [NH3]/[NH4+]
Then convert [OH-] to pOH:
pOH = -log[OH-]
Finally, calculate the pH:
pH = 14 - pOH
ππ»π_3 -> π»^+ + ππ_3^(-)
Since nitric acid is a strong acid, it will completely dissociate in water. Therefore, the initial concentration of π»^+ ions will be equal to the initial concentration of nitric acid, which is 0.175M. The concentration of ππ_3^(-) ions will be zero since it is a spectator ion and does not participate in the formation of additional ππ»π_3 molecules.
[π»^+ ] = 0.175M
[ππ_3^(-) ] = 0M
5. In a 0.175M solution of HCl, HCl is a strong acid that will dissociate completely in water:
π»πΆπ -> π»^+ + πΆπ^(-)
Since HCl is a strong acid, the initial concentration of π»^+ ions will be equal to the initial concentration of HCl, which is 0.175M. The concentration of hydroxide ions (OH-) in this acidic solution will be negligible.
[π»^+ ] = 0.175M
[ππ»^- ] β 0M
6. In a 0.01M solution of Ca(OH)2, Ca(OH)2 is a strong base that will dissociate completely in water:
Ca(OH)2 -> 2ππ»^- + πΆπ^2+
Each Ca(OH)2 molecule dissociates to produce two hydroxide ions (OH-), and the concentration of hydroxide ions will be twice the initial concentration of Ca(OH)2.
[π»^+ ] β 0M
[ππ»^- ] = 2 * 0.01M = 0.02M
7. The ionization constant of water, πΎ_π€, can be expressed as:
πΎ_π€ = [π»_3π^+ ]*[ππ»^- ]
At body temperature, 37Β°C, the value of πΎ_π€ is 1.00 x 10^(-13.60). To determine the concentrations of [π»_3π^+ ] and [ππ»^- ], you would need to calculate the square root of πΎ_π€ since the concentration of [π»_3π^+ ] and [ππ»^- ] will be equal:
[π»_3π^+ ] = [ππ»^- ] = πΎ_π€^(0.5) = (1.00 x 10^(-13.60))^(0.5)
Simply calculate the square root of πΎ_π€ to determine the concentrations at body temperature.
8. To calculate the pH of the following solutions at 25Β°C:
(a) 0.0028M HCl:
Since HCl is a strong acid that dissociates completely, the concentration of H+ ions will be equal to the initial concentration of HCl. Use the formula for pH:
pH = -log[H+]
pH = -log(0.0028)
(b) 0.014M HNO3:
Since HNO3 is also a strong acid that dissociates completely, the concentration of H+ ions will be equal to the initial concentration of HNO3. Use the formula for pH:
pH = -log[H+]
pH = -log(0.014)
(c) 0.00052M HBr:
HBr is a strong acid, so the concentration of H+ ions will be equal to the initial concentration of HBr. Use the formula for pH:
pH = -log[H+]
pH = -log(0.00052)
(d) 0.092M HI:
HI is a strong acid, so the concentration of H+ ions will be equal to the initial concentration of HI. Use the formula for pH:
pH = -log[H+]
pH = -log(0.092)
9. To calculate the pH of a 0.025M solution of ammonia, use the given Kb value:
Kb = [OH-][NH4+]/[NH3]
Since ammonia (NH3) is a weak base, we can assume that the concentration of [OH-] is negligible compared to [NH3]. Therefore, we can simplify the equation to:
Kb β [OH-] * [NH4+]/[NH3]
To determine [OH-] concentration, you can use the equation:
[OH-] = Kb * [NH3]/[NH4+]
Then convert [OH-] to pOH:
pOH = -log[OH-]
Finally, calculate the pH:
pH = 14 - pOH
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