Question

A student dissolves 0.2975 g of the unknown weak acid to give 100. mL solution. S/he obtains the following titration curve for his/her best titration. At the equivalence point, 19.0 mL of 0.1000 M NaOH was added.(a) Calculate the molar mass of the acid.g/mol(b) What is the pKa of the acid?

Answer 1

Answer:

MM = 156.57 g/mol

pH at V = 9.5 mL will be the pKa value for the acid

Explanation:

solution

for the molar mass  MM is here express as

MM = $\frac{mass}{mol}$   .....................1

here  

mass = 0.2975  g

and

mol = MV = 0.1000 × 19 × $10^{-3}$

MV = 0.0019 mol of acid

so

MM =$\frac{0.2975}{0.0019}$  

MM = 156.57 g/mol

and

we know for pKa the easiest way is to find the half equivalence point  is

V half = 0.5 × Vtitration

V half =  0.5 × 19

V half  = 9.5 mL

so  pH at V = 9.5 mL will be the pKa value for the acid