Titration of a 0.85 g sample of unknown soda ash requires 41.0 mLs of 0.120 M HC

Question

Titration of a 0.85 g sample of unknown soda ash requires 41.0 mLs of 0.120 M HCl solution to reach the second equivalence point. Calculate the percent sodium carbonate in the unknown sample. For your information: V_a - 10^+ PH [(1/K_a) (gamma a/gamma) MW of Na_2CO_3 - 106 g/mol] pK_a's of H_2CO_3 - 6.35, 10.33 In plotting the Gran's plot, what is the quantity that corresponds to the y-axis? The x-axis? If for instance, one obtains pH = 9.0 upon addition of 13.0 mLs of 0.1 M HCl titrant to the analyte, what value of x and y would that correspond to in the graph? Titration of a 0.85 g sample of unknown soda ash requires 41.0 mLs of 0.120 M HCl solution to reach the second equivalence point. Calculate the percent sodium carbonate in the unknown sample.

Explanation / Answer

2.

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

1 mole Na2CO3 = 2 mole HCl

NO of mole of HCl consumed = M*V

                            = 41*0.12

                            = 4.92 mmole

so that,

NO of mole of Na2CO3 reacted = 4.92/2 = 2.46 mmole

mass of Na2CO3 reacted = 2.46*10^-3*106 = 0.26 g

% of Na2CO3 in the sample = wt of Na2CO3 / sample *100

                     = 0.26/0.85*100
                    
                     = 30%

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