A 0.4671g sample containing sodium bicarbonate was titirated with hcl requiring 40.72ml. The acid was standardized by titrating 0.1876g of sodium carbonate FW 106mg/mmol requiring 37.86ml of the acid. Find the percentage of nahc03 fw 84mg/mmol in the sample.

My own calculation:

meq hcl = meq na2co3

Na2co3 =106/2
= 53
37.86x=187.6mg/53
37.86x=3.5
x=3.5/37.86
x=0.0935
0.0935 x 40.72 x 84/46.71 x 100 = 68.4%

Who check for me the answer 68.4%

Na2co3 =106/2

= 53
37.86x=187.6mg/53
37.86x=3.5
x=3.5/37.86
x=0.0935
0.0935 x 40.72 x 84/46.71 x 100 = 68.4%
If you multiply this out you get 684.46; the error is you should have divided by 467.1. Then I would not truncate the 68.46% but leave it to four significant figures.

Who check for me the answer 68.4%

Well, you know what they say - two heads are better than one, unless both heads are clowns! But let's take a closer look at your calculations.

First, we need to calculate the Meq (milliequivalents) of Na2CO3:

Meq Na2CO3 = (0.1876g / 106mg/mmol) / 2
= 0.00176 mol / 2
= 0.00088 Meq

Next, we can calculate the Meq of HCl used in the titration:

Meq HCl = 37.86ml * Meq Na2CO3 / 40.72ml
= 37.86ml * 0.00088 Meq / 40.72ml
= 0.000825 Meq

Now, we can calculate the Meq of NaHCO3 in the sample:

Meq NaHCO3 = Meq HCl

Finally, we can calculate the percentage of NaHCO3 in the sample:

Percentage NaHCO3 = (Meq NaHCO3 * FW of NaHCO3) / Sample weight * 100
= (0.000825 Meq * 84mg/mmol) / 0.4671g * 100
≈ 18.7%

So, the correct answer is approximately 18.7%, not 68.4%.

To verify the calculation, let's break it down step-by-step:

First, let's find the milliequivalents (meq) of HCl used to titrate the sodium carbonate:

meq HCl = (0.1876g sodium carbonate) / (106mg/mmol sodium carbonate) = 0.00177 mol sodium carbonate

Next, calculate the percentage concentration of the HCl solution:

%HCl = (meq HCl * 1000 mL) / volume of HCl used
= (0.00177 mol * 1000 mL) / 37.86 mL
= 46.76 mol/L

Now, let's find the milliequivalents of HCl used to titrate the sodium bicarbonate:

meq HCl = (40.72 mL) * (46.76 mol/L)
= 1906.67 meq

Since meq HCl = meq NaHCO3, we have:

meq NaHCO3 = 1906.67 meq

Next, let's find the milligrams of NaHCO3 in the sample:

mg NaHCO3 = (1906.67 meq * 84 mg/mmol * 1 mmol/1000 mg) = 160.2 mg

Finally, calculate the percentage of NaHCO3 in the sample:

% NaHCO3 = (mg of NaHCO3 / mass of the sample) * 100
= (160.2 mg / 0.4671 g) * 100
= 343.48%

Therefore, the correct answer is 343.48%, not 68.4%.

To verify your calculation, let's break down the steps and check each calculation:

1. Calculate the milliequivalents (meq) of HCl used to titrate the sodium carbonate:
meq(HCl) = meq(Na2CO3)
meq(Na2CO3) = (0.1876g / 106mg/mmol) * 2
meq(Na2CO3) = 0.00353 mol

2. Calculate the milliequivalents (meq) of NaHCO3 in the sample:
meq(NaHCO3) = meq(HCl) = 0.00353 mol

3. Calculate the mass of NaHCO3 in the sample:
mass(NaHCO3) = (meq(NaHCO3) * 84mg/mmol) / 1000
mass(NaHCO3) = (0.00353 mol * 84mg/mmol) / 1000
mass(NaHCO3) = 0.29652g

4. Calculate the percentage of NaHCO3 in the sample:
% NaHCO3 = (mass(NaHCO3) / sample mass) * 100
% NaHCO3 = (0.29652g / 0.4671g) * 100
% NaHCO3 = 63.51%

Therefore, the percentage of NaHCO3 (sodium bicarbonate) in the sample is approximately 63.51%, which is different from the answer you provided.

Please double-check your calculation steps or provide any additional information if needed.