# Find the pH of a solutino prepared by mixing 40.0 mL of a .02 M HCl solution with 200.0 mL of a .20 M HCN solution. Assume volumes to be additive. Ka for HCN= 1.0 x 10^-10. The answer is 2.4,...

Find the pH of a solutino prepared by mixing 40.0 mL of a .02 M HCl solution with 200.0 mL of a .20 M HCN solution.

Assume volumes to be additive. Ka for HCN= 1.0 x 10^-10.

The answer is 2.4, but I have no idea how to do it.

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`HCl rarr H^++Cl^-`

`HCN harr H^++CN^-`

So we get `H^+` for the solution from both HCl and HCN.

The `K_a` of HCN is a very small value. So we cannot expect that we have get considerable amount of `H^+`from HCN. On the other hand according to the Le Chatelier's principal since we have `H^+` from HCl the forward reaction of HCN to give `H^+` will be decreased.

So in the final solution the H^+ from HCN is negligible.

Only `H^+` from HCl will account in PH in the solution.

Amount of H^+ from HCl `= 0.02/1000xx40 = 0.0008mol`

When we mix the solutions the final volume is 240ml. So there will be 0.0008moles of` H^+` in the final solution.

`[H^+] = 0.0008/240xx1000 = 0.00333M`

`PH = -log[H^+]`

`PH = -log(0.00333)`

`PH = -log(3.33xx10^(-3))`

`PH = 3-log.3.33`

`PH = 3-0.522`

`PH = 2.478`

*So the PH of the final solution is 2.478*