Question

pH of `0.1(M)` `NaH_2PO_4` is what?

Answer 1

pH = 4.10

Explanation:

If it was a strong acid then the concentration of `H^+` that dissociates from `NaH_2PO_4` would be `2 xx 0.1 = 0.2M`

But `NaH_2PO_4` is a weak acid. It will dissociated partially.

If `x` represents concentration of acid that dissociates then

`x = sqrt(K_a xx C)`
(see Ernest answer for more details about this formula)

In order to find pH of a weak acid we should know acid dissociation constant (`K_a`) value.

`K_a` of `NaH_2PO_4` is `6.2 xx 10^-8`

`x = sqrt(6.2 xx 10^-8 xx 0.1)`

`x = 7.9 xx 10^-5 M`

`pH = -log[H^+]`

`pH = -log[7.9 xx 10^-5]`

`pH = 4.10`

Answer 2

`"pH = 4.10"`

Explanation:

`"NaH"_2"PO"_4` dissociates completely in solution:

`"NaH"_2"PO"_4 → "Na"^"+" + "H"_2"PO"_4^"-"`

The dihydrogen phosphate ion is a weak acid:

`"H"_2"PO"_4^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HPO"_4^"2-"; K_text(a) = 6.23 × 10^"-8"`

We can use an ICE table to calculate the concentrations of the ions in solution.

Let's rewrite the equation as

`color(white)(mmmmmmmm)"HA"^"-" +color(white)(ll) "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"2-"`
`"I/mol·L"^"-1":color(white)(mmm)0.1color(white)(mmmmmmmll)0color(white)(mmm)0`
`"C/mol·L"^"-1":color(white)(mmm)"-"xcolor(white)(mmmmmmml)"+"xcolor(white)(mm)"+"x`
`"E/mol·L"^"-1":color(white)(mm)"0.1 -"xcolor(white)(mmmmmmm)xcolor(white)(mmm)x`

`K_text(a) = (["H"_3"O"^"+"]["A"^"2-"])/(["HA"]) = (x × x)/("0.1 -"color(white)(l)x) = x^2/("0.1 -"color(white)(l)x) = 6.23 × 10^"-8"`

Check for negligibility:

`0.1/(6.23 × 10^"-8") = 2 × 10^6 ≫ 400`.

∴ `x ≪ 0.1`.

Then

`x^2/0.1 = 6.23 × 10^"-8"`

`x^2 = 0.1 × 6.23 × 10^"-8" = 6.2 × 10^"-9"`

`x = 7.9 × 10^"-5"`

`["H"_3"O"^"+"] = x color(white)(l)"mol/L" = 7.9 × 10^"-5"color(white)(l)"mol/L"`

`"pH" = "-log"["H"_3"O"^"+"] = "-log"(7.9 × 10^"-5") = 4.10`