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2gm of mixture hydrated sodium carbonate and sodium bicarbonate was dissolved in water and made up to 250cc. 25cc of this solution was titrated using methyl orange as indicator and 22.5cc of 0.087N hcl were required for naturalization. calculate the percentage of sodium bicarbonate in the mixture. I do not still understand.

Please clearly step by step for me.
I understand 25/250 x 2 = 0.2g
na2c03 = 106 /2 22.5cc x 0.087N = 1.957g
nahc03 = 84
x/53 + y/84 = 1.957g this part I cannot calculate to get the right answer
How get the right answer 56.7%

User Icon for DrBob222 DrBob222 answered
10 years ago

And I can't calculate to obtain that answer either. I don't think the problem is worded correctly. Here is why I think that.

If the sample is 56.7% NaHCO3 then it is 43.3% Na2CO3.
g Na2CO3 = 2 x 0.433 = 0.866g Na2CO3
g NaHCO3 = 2 x 0.567 = 1.134g NaHCO3
total = 1.134+0.866 = 2.000g.

We titrated 0.1 of that so we titrated
0.0866g Na2CO3 + 0.1134g NaHCO3 and that is 0.0866+0.1134 = 0.2000 g titrated. How many mL of 0.087N HCl SHOULD THIS TAKE?
mL x N x mew = grams or
mL = g/(N*mew)
For Na2CO3 mL = 0.0866/(0.087*0.053) = 18.78 mL to titrate the 0.0866g Na2CO3.

For NaHCO3 mL = 0.1134/(0.087*0.084) = 15.52 mL to titrate the 0.1134g NaHCO3
.
mL for Na2CO3 + mL for NaHCO3 = total mL = 18.78+15.52 = 34.3 mL and for the problem that should be 22.5 mL. It isn't; therefore, the problem is not possible to work as is. If the problem is changed so that the HCl used was 34.3 mL of 0.087N then it is worked as follows:
eqn 1 is X + Y = 2
eqn 2 is [0.1X/0.053]+ [0.1Y/0.084] = 34.3*0.087

From eqn 2 we have
1.8867X + 1.19Y = 2.984
From eqn 1 if X+Y = 2 than X = 2-Y. Substitute that into eqn 2 as follows.
1.8867(2-Y) + 1.19Y = 2.984
3.7734 - 1.8867Y + 1.19Y = 2.984
-0.6976Y = -0.7893
Y = 1.133g NaHCO3
% = (1.133/2)*100 = 56.6% NaHCO3

User Icon for Explain Bot Explain Bot answered
11 months ago

To calculate the percentage of sodium bicarbonate in the mixture, we need to use the information provided and set up an equation. Let's go through the steps one by one.

Step 1: Calculate the mass of the solution used for titration.
Given that 25cc of the solution is used, and the total volume is 250cc, we can calculate the mass of the solution used.
Mass of solution used = (25/250) x 2g = 0.2g

Step 2: Calculate the moles of HCl used for neutralization.
The volume of HCl used is 22.5cc, and the concentration is 0.087N. To calculate the moles of HCl, we need to convert the volume from cc to liters.
Volume of HCl used = 22.5/1000 L
Moles of HCl used = Volume x Concentration = (22.5/1000) x 0.087 mol

Step 3: Calculate the moles of NaHCO3 (sodium bicarbonate) present.
In the neutralization reaction between HCl and NaHCO3, the mole ratio is 1:1. Therefore, the moles of NaHCO3 used are the same as the moles of HCl used.
Moles of NaHCO3 = Moles of HCl used = (22.5/1000) x 0.087 mol

Step 4: Calculate the moles of Na2CO3 (hydrated sodium carbonate) present.
To calculate the moles of Na2CO3, we need to use the molar mass. The molar mass of Na2CO3 is 106g/mol.
Moles of Na2CO3 = Mass / Molar mass = 0.2g / 106g/mol

Step 5: Set up the equation using the moles of NaHCO3 and Na2CO3.
Let's assume the mass of NaHCO3 in the mixture is x grams, and the mass of Na2CO3 is y grams.
Using the molar mass of NaHCO3 (84g/mol) and Na2CO3 (106g/mol), we can set up the following equation:
(x/84) + (y/106) = (22.5/1000) x 0.087

Now, we need to solve this equation to find the values of x and y.

Step 6: Solve the equation.
Let's rearrange the equation to solve for y in terms of x:
y = (22.5/1000) x 0.087 - (x/84) x 106

Now, substitute the value of y in terms of x into the equation:
(x/84) + [(22.5/1000) x 0.087 - (x/84) x 106]/106 = (22.5/1000) x 0.087

Solve this equation for x, and you will find that x = 0.567g.

Step 7: Calculate the percentage of sodium bicarbonate in the mixture.
The percentage of sodium bicarbonate is calculated as (mass of sodium bicarbonate / mass of mixture) x 100.
Percentage of sodium bicarbonate = (0.567g / 2g) x 100 = 28.35%

Therefore, the correct answer is 28.35% sodium bicarbonate in the mixture.