You have a 0.4674 g sample of sodium carbonate (Na2CO3), which you suspect is co

Question

You have a 0.4674 g sample of sodium carbonate (Na2CO3), which you suspect is contaminat-ed with an inert white substance. You place this entire mass into a 250 mL Erlenmeyer flask, and then decide to titrate it against standardized 0.1755 M hydrochloric acid, which you fill a 100 mL buret with.

a) Write a balanced chemical equation for this process below:

b) Upon completion of the titration, the following data is collected: initial buret reading = 0.03 mL; final buret reading = 50.29 mL. Based on this data,determine the % purity of the sodium carbonate sample

Explanation / Answer

m = 0.4674 Na2CO3

MW = 105.9888

mol = mass/MW = 0.4674/105.9888 = 0.00440989991

M = 0.1755 M of HCl

a)

2HCl + Na2CO3 -> 2NaCl + H2O + CO2

b)

V = 0.03 initial

Vf = 50.29

VT = 100

then Vt = 50.29 - 0.03 = (100-50.26) ml used = 49.74

mol of acid used = M*V = 49.74*0.1755 =8.72937mmol of acid

then there are

8.72937/2 mmol of Na2CO3

4.364685 mmol of Na2CO3

mass = (4.364685*10^-3)(105.9888) = 0.462607

% purity = 0.462607/0.4674 *100 = 98.9745%

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