Question

A 0.050 M solution of a weak monoprotic acid, HA, has [H₃O⁺] = 3.8 x 10^-5. What is the value of K_a for this acid?

2.6 x 10^-11

2.9 x 10^-8

7.5 x 10^-3

7.0 x 10^-8

7.0 x 10^-7

Answer 1

We know that - ÞH = - dog [H3O+] Given - Ho+1 = 3.8x10-5. .: ÞH = -loq [H30+ ] = - dog (3.8 x 10-5) = 4:42 :ÞH of solution in 4.42 . Also we know that the bit of a monoprotic weak acid (HA) a in- pH = Ź pka - I loqe ; kas ionination constant ' c = concentration of 1 weak steid

Criven — сО. ОБО М pH = 4.42 (calculated) Ka=? Accordingly, A pH = /2 bka - Ž soge or, pH + h log c = Ž pka or, 2x (PH + 1/2 loge) = pka or, 2xbH + Loge = -loq ka or, (2 x 4:42) + 10% (0.050) = - dog ka or, 8.84 +(- 1:30) = - 40q ka or, - log Ka = 8,84 - 4:30 or - log ka = 7.54 or, 10g Ka = -7:54 or, Ka = 10-7-54 = 2.88x10-8 ~ 2.9 x10-8 Ams.

... The value of Ka for the acid will be 2.9810-8. Am