### Given the pK of a Salt, Calculate the pH of a Solution of the Acid or Base

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Typically, when doing this type of problem, you usually see one of two things:

1) You are given the concentration and K_{a} (or pK_{a}) of an acid, then asked to calculate the pH.

2) You are given the concentration and K_{b} (or pK_{b}) of a base, then asked to calculate the pH.

In the examples below, you are given (a) the pK of a salt of an acid or base and (b) the concentration, then asked to calculate the pH. There is just one additional step when compared to the technique needed for (1) and (2) just above.

**Problem #1:** What is the pH of a 0.262 M NH_{3} solution? The pK_{a} for NH_{4}^{+} = 9.248.

**Solution:**

Determine the K_{b} for ammonia:

Since K_{a}K_{b} = K_{w} = 1.00 x 10^{-14}, we know:
pK_{a} + pK_{b} = 14.000
9.248 + x = 14.000

x = 4.752

Then:

K_{b} = 10^{-pKb}
K_{b} = 10^{-4.752}

K_{b} = 1.77 x 10^{-5}

The step above is the additional step I referred to above. Now, the problem becomes this:

What is the pH of a 0.262 M NH_{3} solution? The K_{b} for NH_{3} = 1.77 x 10^{-5}.

You can see the rest of the solution technique here. The answer to the above problem is 11.333

**Problem #2:** What is the pH of a 0.262 M HCN solution? The pK_{b} for CN¯ = 4.788.

**Solution:**

Determine the K_{a} for HCN:

Since K_{a}K_{b} = K_{w} = 1.00 x 10^{-14}, we know:
pK_{a} + pK_{b} = 14.000
x + 4.788 = 14.000

x = 9.212

Then:

K_{a} = 10^{-pKa}
K_{a} = 10^{-9.212}

K_{a} = 6.14 x 10^{-10}

The step above is the additional step I referred to above. Now, the problem becomes this:

What is the pH of a 0.262 M HCN solution? The K_{a} for HCN = 6.14 x 10^{-10}.

You can see the rest of the solution technique here. The answer to the above problem is 4.897

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