Question

A 25.0 ml sample of 0.150 M nitrous acid is titrated with a0.150 M NaOH solution. What is the pH at the equivalence point? theKa of Nitrous Acid is 4.50 x 10-4.

Answer 1

See the answer by @Andrea. https://socratic.org/users/andrea-b-3

Answer 2

pH= 8.11

Explanation:

When the titration occurrs you have used 25,0 mL of NaOH so you in total have 50.0 mL of `NaNO_2` that (because the volume is double) will be 0.075 M
pH of equivalent point is the pH of this solution of `NaNO_2` 0.075 M
It is a salt with a basic hydrolysis
`[OH^(-)]= sqrt (K_w/K_a xx C_s)=sqrt((1 xx 10^(-14))/(4.5 xx 10^(-4)) xx 7.5 xx 10^(-2) )= sqrt(1,67 xx 10^(-12))= 1.29 xx 10^-6`
pOH = 5.89
pH= 8.11

Answer 3

You should treat this problem with care... most students would `(i)` use the Henderson-Hasselbalch equation and use `"HNO"_2` as the acid and `"NaOH"` as the base, `(ii)` use the `K_a` and thus assume `"HNO"_2` is dissociating in water, or `(iii)` forget about dilution and use `"0.150 M" = ["NO"_2^(-)]_i`. That doesn't make sense.

The `"pH"` is `8.11`.

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By neutralizing to the equivalence point, only `"NO"_2^-` remains. That comes from

`"0.150 mol HNO"_2/"L" xx "0.025 L" = "0.00375 mols NO"_2^(-)` formed

upon exact neutralization of `"HNO"_2` with `"NaOH"`. This is now in a volume of `"25.0 mL" + "25.0 mL" = "50.0 mL" = "0.050 L"`.

Therefore, we get a diluted concentration of

`["NO"_2^-]_i = "0.00375 mols"/"0.050 L" = "0.075 M NO"_2^-`,

as expected since the volume was doubled (so the concentration should halve).

Now, this nitrite will associate in an equilibrium within water:

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

`"I"" "0.075" "" "" "" "-" "" "0" "" "" "" "0`
`"C"" "-x" "" "" "" "-" "" "+x" "" "+x`
`"E"" "0.075-x" "" "-" "" "x" "" "" "" "x`

Here we have a BASE, and so we must use the `K_b`:

`K_aK_b = K_w`

`=> K_w/K_a = K_b = 10^(-14)/(4.50 xx 10^(-4)) = 2.22 xx 10^(-11)`

It is apparent that this `K_b` is very small, so the small `x` approximation works here very well. This is true usually when `K` is on the order of `10^(-5)` or less.

Write the mass action expression:

`K_b = 2.22 xx 10^(-11) = x^2/(0.075 - x) ~~ x^2/0.075`

Hence,

`x -= ["OH"^(-)] = sqrt(0.075K_b)`

`= sqrt(0.075 cdot 2.22 xx 10^(-11))`

`= 1.29 xx 10^(-6) "M OH"^-`

Therefore,

`"pOH" = -log["OH"^(-)] = 5.89`

`color(blue)("pH") = 14 - "pOH" = color(blue)(8.11)`