Question

What is the pH of `3.1 times 10^-3` `M` `HCl`?

Answer 1

There are two ways you can do this. The easy way is to realize that `"HCl"` is a strong acid, so its dissociation is considered complete, and `["HCl"] = ["H"^(+)]`.

EASY WAY

Recall:

`"pH" = -log["H"^(+)]`

From the knowledge that `"pH" = -log["H"^(+)] = -log["HCl"]`, we can say:

`color(blue)("pH") = -log(3.1 xx 10^(-3) "M") = 2.50_(8638306)`

`= color(blue)(2.51)`

HARD WAY

The hard way is to look up the `K_a` of `"HCl"` and find that it is around `1.3 xx 10^6` and do an ICE Table as if it were an actually not-heavily-skewed equilibrium. I would never advise any chemistry student to do this...

`"HCl"(aq) " "" " -> " "" ""H"^(+)(aq) + "Cl"^(-)(aq)`

`"I"" "3.1 xx 10^(-3)"M"" "" "" "" ""0 M"" "" ""0 M"`
`"C"" "-x" "" "" "" "" "" "" "+x" "" "+x`
`"E"" "(3.1 xx 10^(-3) -x)"M"" "" "x" "" "" "x`

`K_a = 1.3 xx 10^6 = x^2/(3.1 xx 10^(-3) - x)`

We could simply make the approximation that `3.1 xx 10^(-3) ~~ x`, but we are supposing we don't know that yet. So, let's just solve for the quadratic equation.

`x^2 + K_ax - 3.1 xx 10^(-3)K_a = 0`

The quadratic formula gives, with `a = 1`, `b = K_a`, `c = -3.1 xx 10^(-3)K_a`:

`x = 0.003099999993`,

and we reject the root `x ~~ -1.3 xx 10^6` since `["HCl"]_(eq)` would have become `3.1 xx 10^(-3) - (-1.3 xx 10^6) ~~ 1.3 xx 10^6`, which would falsely imply that `"HCl"` doesn't dissociate at all.

Thus, we get that

`K_a = (3.099999993 xx 10^(-3))^2/(3.1 xx 10^(-3) - 3.099999993 xx 10^(-3))`

`= 1.37 xx 10^6 ~~ 1.3 xx 10^6`,

as it was when we started, and

`["HCl"]_(eq) = 3.1 xx 10^(-3) - 3.099999993 xx 10^(-3) ~~ 0`,

and

`["H"^(+)]_(eq) = 3.099999993 xx 10^(-3)` `"M"` `~~` `3.1 xx 10^(-3)` `"M"`,

as expected. Finding `K_a` is still valid, however, since the denominator isn't exactly zero... So, if we really wanted the "exact" answer:

`"pH" = -log["H"^(+)] = 2.50_(8638307) ~~ 2.51`

as before, only differing past the 8th decimal place (so who cares?).