A 5.329 g sample of a solid, weak, monoprotic acid is used to make a 100.0 mL so

Question

A 5.329 g sample of a solid, weak, monoprotic acid is used to make a 100.0 mL solution. 34.00 mL of the resulting acid solution is then titrated with 0.09506 M NaOH. The pH after the addition of 21.00 mL of the base is 5.40, and the endpoint is reached after the addition of 47.78 mL of the base.

(a) How many moles of acid were present in the 34.00 mL sample?

(b) What is the molar mass of the acid?

(c) What is the pKa of the acid?

I figured out a and c and got:

a. .0045

c. 5.51

I can't figure out part B

Explanation / Answer

millimoles of acid = 34 x M

millimoles of base = 0.09506 x21= 1.99626

HA       +      NaOH ---------------------> NaA   + H2O    

34 M    1.99626 1.99626   

it is acidic buffer

Henderson-Hasselbalch equation

pH = pKa + log[salt/acid]

pH = 5.40

pKa = 5.51 (your calculated value )

5.40 = 5.51 + log[1.99626 /34M]

-0.11 = log[1.99626 /34M]

0.776 = 1.99626 /34M

34 M = 2.57

M = 2.57/34

   = 0.0755

molarity = (W/MW) x 1000/V(ml)

W= 5.329g

V= 100ml

0.0755 = (5.329/MW) x 1000/100

Mw = 705.8 g/mol

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