Question

What is the pH at the equivalence point of the titration of 30 mL of 0.2 M nitrous acid with 0.1 M NaOH at 25 °C?

A. 8.1
B. 8.3
C. 7.0
D. 9.1
E. 3.3
The correct answer is A, but how?

Answer 1

If you got `B`, you did not account for dilution.

The question showed:

A. 8.1
B. 8.3
C. 7.0
D. 9.1
E. 3.3
The correct answer is A, but how?

Explanation:

The concentration of `"0.1 M"` `"NaOH"` is half the concentration of the monoprotic `"HNO"_2`, so twice the volume is needed, i.e. `"60 mL"` of `"NaOH"`. At the equivalence point, only `"NO"_2^(-)` is left.

Hence, the pH must be slightly basic, and this calculation requires that you give the `K_a`... The `K_a` of `"HNO"_2` MIGHT be `4.5 xx 10^(-4)`, but you'll have to verify that.

Thus, at `25^@ "C"`,

`K_b = K_w/K_a = 10^(-14)/(4.5 xx 10^(-4)) = 2.22 xx 10^(-11)`

You should be able to explain why we must use `K_b`, the base association constant, for

`"NO"_2^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HNO"_2(aq) + "OH"^(-)(aq)`

The concentration of `"NO"_2^(-)` was diluted by a factor of

`("30 mL HNO"_2 + "60 mL NaOH")/"30 mL HNO"_2 = 3`,

i.e. `"0.200 M HNO"_2` before mixing `->` `"0.066 M NO"_2^(-)` after mixing.

For this the mass action expression becomes:

`K_b = (["HNO"_2]["OH"^(-)])/(["NO"_2^(-)])`

`= x^2/(0.066 - x)`

We can clearly see that `K_b` is very small, so `x "<<" "0.200 M"`. Therefore,

`2.22 xx 10^(-11) = x^2/0.066`

`=> color(blue)(x = 1.21 xx 10^(6) "M")`

Knowing what `x` represents (which is?), you should then fill in the steps to get that the `"pH"` is `8.08`.