# pH of 0.1(M) NaH_2PO_4 is what?

Aug 8, 2017

pH = 4.10

#### Explanation:

If it was a strong acid then the concentration of ${H}^{+}$ that dissociates from $N a {H}_{2} P {O}_{4}$ would be $2 \times 0.1 = 0.2 M$

But $N a {H}_{2} P {O}_{4}$ is a weak acid. It will dissociated partially.

If $x$ represents concentration of acid that dissociates then

$x = \sqrt{{K}_{a} \times C}$

In order to find pH of a weak acid we should know acid dissociation constant (${K}_{a}$) value.

${K}_{a}$ of $N a {H}_{2} P {O}_{4}$ is $6.2 \times {10}^{-} 8$

$x = \sqrt{6.2 \times {10}^{-} 8 \times 0.1}$

$x = 7.9 \times {10}^{-} 5 M$

$p H = - \log \left[{H}^{+}\right]$

$p H = - \log \left[7.9 \times {10}^{-} 5\right]$

$p H = 4.10$

Aug 9, 2017

$\text{pH = 4.10}$

#### Explanation:

${\text{NaH"_2"PO}}_{4}$ dissociates completely in solution:

$\text{NaH"_2"PO"_4 → "Na"^"+" + "H"_2"PO"_4^"-}$

The dihydrogen phosphate ion is a weak acid:

$\text{H"_2"PO"_4^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HPO"_4^"2-"; K_text(a) = 6.23 × 10^"-8}$

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

Let's rewrite the equation as

$\textcolor{w h i t e}{m m m m m m m m} \text{HA"^"-" +color(white)(ll) "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"2-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m m} 0.1 \textcolor{w h i t e}{m m m m m m m l l} 0 \textcolor{w h i t e}{m m m} 0$
$\text{C/mol·L"^"-1":color(white)(mmm)"-"xcolor(white)(mmmmmmml)"+"xcolor(white)(mm)"+} x$
$\text{E/mol·L"^"-1":color(white)(mm)"0.1 -} x \textcolor{w h i t e}{m m m m m m m} x \textcolor{w h i t e}{m m m} x$

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