Question

Calculate the pH of the solution resulting from the addition of 10.0 mL of 0.10 M NaOH to 50.0 mL of a 0.10 M solution of aspirin (acetylsalicic acid, K_a = 3.0 × 10^–4) solution.

	A.	2.9
	B.	10.5
	C.	4.1
	D.	3.5
	E.	1.8

Answer 1

d. 3.5

given data

10 ml - 0.10 M NaOH                                                     50 ml-0.10 M aspirin (C9H8O4

10 ml= 0.01 L                                                                 50ml= 0.05L

0.01 x 0.10 /1                                                                 0.05 x 0.10/1

= 0.001 mol NaOH or OH^{-                                                                =}0.005 mol

C₉H₈O_4(s)+NaOH_(aq)→C₉H₇O_4(s)^-Na⁺+H₂O_(l)

0.005/0.06 =0.083 M

pH=pKa + log( C₉H₇O4 ^-/ HC₉H₇O4

pKa= -log Ka

= -log 3.0 X 10-4

=3.5

=3.5+ log (0.083/0.083)

=3.5