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# What is the pHof a weak monoprotic acid in a 0.01482M solution, given its pka=4.20?

### 2 Answers

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- BobbyLv 72 months ago
for a weak acid [H=] = sqrt ( Ka *c)

this assumes a very small dissociation of the weak acid

this is reasonable given the very small Ka

so there is no need for ICE tables

Ka = 10^-4.2 = 6.310 * 10 ^-5

[H+] = sqrt( 6.310 *10 ^-5 * 0.01482) =9.67E-04

pH = log(H+) = -log(9.67E-04 )

= 3.01

there is your answer

- jacob sLv 72 months ago
0.01482 =[HA]

pka=4.20 ; Ka = 6.309 x 10^-5

HA ⇔ H+ + A -

I 0.01482. - -

C -y +y +y

E 0.01482 -y. y y

.: Ka = [H+][A-]/[HA] = y²/(0.01482 -y) = 6.309 x 10^-5

y= 0.000935 = [H+]

pH = - log[H+]= - log[0.000935]= 3.03

Please rate Thank you.

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