Question

If the ?aof a monoprotic weak acid is 7.5×10−6, what is the pH of a 0.21 Msolution of this acid?

PLEASE BREAK IT DOWN STEP BY STEP

Answer 1

K_a=7.5×10^-6(given)

Concentration=0.21M.Assume HA is the weak monoprotic acid.

HA<==> H⁺ + A^- . Ka = 7.5×10^-6
0.21. 0 ......0 . initial
-x .........+x ... +x ................. change
0.21-x ... x ..... x .................. equilibrium

Ka = [H⁺][A^-] / [HA] =(x×x)/(0.21-x)
7.5x10^-6= x² / (0.21-x)

As x is very small than 0.21.we can take 0.21-x as 0.21.

X²=0.21×7.5×10^-6=1.575×10^-6

X=√1.575×10^-6=1.2×10^-3

Therefore x=[H⁺]=1.2×10^-3

pH=-log[H+]=-log(1.2×10^-3)=2.9

pH is 2.9

Thank you