The pH of a 0.25 M weak acid is 2.036. What is the Ka of the...

Question

Question

The pH of a 0.25 M weak acid is 2.036. What is the K_a of the acid? HA(aq) + H₂O (l) ↔ H₃O⁺(aq) + A^-(aq)
1. 3.5 X 10^-4
2. 9.2X 10^-3
3. 7.0 X10^-5
4. 5.3 X 10^-2
5. 6.3 X 10^-7

Explanation:

[ H₃O⁺] = antilog (-pH) =10^-pH =10^-2.306=9.2 X10^-3 M

[ H₃O⁺] = [A^-]=9.2 X10^-3 M

[HA] = 0.25 M- 0.0092 M=0.241 M

K_a = [9.2 X10^-3 M] [9.2 X10^-3 M]/0.241 =3.52X10^-4

^{I don't get why
10^-2.036= 9.2x10^3}

science chemistry

Answer 1

Answer #1

Answer 2

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