Question
A student begins with 25 mL of a 0.434 M solution of HI and slowly adds a solution of 0.365 M NaOH.
1. What is the pH after 15.00 ml of NaOH solution has been added?
2. What is the pH after 40 ml of NaOH solution has been added?
1. What is the pH after 15.00 ml of NaOH solution has been added?
2. What is the pH after 40 ml of NaOH solution has been added?
Answers
DrBob222
You must know where the equivalence point is in order to know where you are on the titration curve.
mL acid x M acid = mL base x M base.
25*0.434 = mL x 0.365
mL base = estimated 30 mL so 15 mL is BEFORE the eq point and 40 mL is AFTER the eq point.
A. mols HI = M x L = ?
mols NaOH = M x L (0.015 x 0.365) = ?
Subtract, giving you mols HI, divide by total L (15 mL + 25 mL) to give M and convert to pH.
B. mols HI = M x L
mols NaOH= M x L
Subtract giving mols NaOH and divide by total L for M, then convert to pOH and pH.
mL acid x M acid = mL base x M base.
25*0.434 = mL x 0.365
mL base = estimated 30 mL so 15 mL is BEFORE the eq point and 40 mL is AFTER the eq point.
A. mols HI = M x L = ?
mols NaOH = M x L (0.015 x 0.365) = ?
Subtract, giving you mols HI, divide by total L (15 mL + 25 mL) to give M and convert to pH.
B. mols HI = M x L
mols NaOH= M x L
Subtract giving mols NaOH and divide by total L for M, then convert to pOH and pH.
bonjo
1.mole of NaOH added = cv = 0.365x15e-3 = 5.48e-3 moles.
Reaction;
HI + NaOH --> NaI + H2O
pH = -log[H+]
the [H+] is the concentration of the acid as 15ml NaOH has been added.
the equation states that, 1 mole of NaOH neutralize 1 mole of HI. So, after adding 15ml, we introduce 5.48e-3 moles of NaOH in the solution. So, this mole should also neutralize 5.48e-3moles of HI.
concentration is moles/volume. The volume of acid is 25mL. with that, use -log[H+].
2. again, find the mole for 40ml of NaOH to find the mole of HI been neutralized. then use that mole to calculate the concentration..
hope that helpss..
Reaction;
HI + NaOH --> NaI + H2O
pH = -log[H+]
the [H+] is the concentration of the acid as 15ml NaOH has been added.
the equation states that, 1 mole of NaOH neutralize 1 mole of HI. So, after adding 15ml, we introduce 5.48e-3 moles of NaOH in the solution. So, this mole should also neutralize 5.48e-3moles of HI.
concentration is moles/volume. The volume of acid is 25mL. with that, use -log[H+].
2. again, find the mole for 40ml of NaOH to find the mole of HI been neutralized. then use that mole to calculate the concentration..
hope that helpss..