pH of $0.1 (M)$ $0.1(M)$ $N a H_{2} P O_{4}$ $NaH_2PO_4$ is what?

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pH of $0.1 (M)$ $0.1(M)$ $N a H_{2} P O_{4}$ $NaH_2PO_4$ is what?

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Answer 1 · 2017-08-08T13:00:36

pH = 4.10

Explanation:

If it was a strong acid then the concentration of $H^{+}$ $H^+$ that dissociates from $N a H_{2} P O_{4}$ $NaH_2PO_4$ would be $2 \times 0.1 = 0.2 M$ $2 xx 0.1 = 0.2M$

But $N a H_{2} P O_{4}$ $NaH_2PO_4$ is a weak acid. It will dissociated partially.

If $x$ $x$ represents concentration of acid that dissociates then

$x = \sqrt{K_{a} \times C}$ $x = sqrt(K_a xx C)$
(see Ernest answer for more details about this formula)

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

$K_{a}$ $K_a$ of $N a H_{2} P O_{4}$ $NaH_2PO_4$ is $6.2 \times 10^{- 8}$ $6.2 xx 10^-8$

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

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

$p H = - log [H^{+}]$ $pH = -log[H^+]$

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

$p H = 4.10$ $pH = 4.10$

Answer 2 · 2017-08-09T00:11:53

$pH = 4.10$ $"pH = 4.10"$

Explanation:

${NaH}_{2} {PO}_{4}$ $"NaH"_2"PO"_4$ dissociates completely in solution:

${NaH}_{2} {PO}_{4} \to {Na}^{+} + H_{2} {PO}_{4}^{-}$ $"NaH"_2"PO"_4 → "Na"^"+" + "H"_2"PO"_4^"-"$

The dihydrogen phosphate ion is a weak acid:

$H_{2} {PO}_{4}^{-} + H_{2} O ⇌ H_{3} O^{+} + {HPO}_{4}^{2-}; K_{a} = 6.23 \times 10^{-8}$ $"H"_2"PO"_4^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HPO"_4^"2-"; K_text(a) = 6.23 × 10^"-8"$

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

Let's rewrite the equation as

$m m m m m m m m {HA}^{-} + l l H_{2} O ⇌ H_{3} O^{+} + A^{2-}$ $color(white)(mmmmmmmm)"HA"^"-" +color(white)(ll) "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"2-"$
${I/mol\cdotL}^{-1} : m m m 0.1 m m m m m m m l l 0 m m m 0$ $"I/mol·L"^"-1":color(white)(mmm)0.1color(white)(mmmmmmmll)0color(white)(mmm)0$
${C/mol\cdotL}^{-1} : m m m - x m m m m m m m l + x m m + x$ $"C/mol·L"^"-1":color(white)(mmm)"-"xcolor(white)(mmmmmmml)"+"xcolor(white)(mm)"+"x$
${E/mol\cdotL}^{-1} : m m 0.1 - x m m m m m m m x m m m x$ $"E/mol·L"^"-1":color(white)(mm)"0.1 -"xcolor(white)(mmmmmmm)xcolor(white)(mmm)x$

$K_{a} = \frac{[H_{3} O^{+}] [A^{2-}]}{[HA]} = \frac{x \times x}{0.1 - l x} = \frac{x^{2}}{0.1 - l x} = 6.23 \times 10^{-8}$ $K_text(a) = (["H"_3"O"^"+"]["A"^"2-"])/(["HA"]) = (x × x)/("0.1 -"color(white)(l)x) = x^2/("0.1 -"color(white)(l)x) = 6.23 × 10^"-8"$

Check for negligibility:

$\frac{0.1}{6.23 \times 10^{-8}} = 2 \times 10^{6} ≫ 400$ $0.1/(6.23 × 10^"-8") = 2 × 10^6 ≫ 400$ .

∴ $x ≪ 0.1$ $x ≪ 0.1$ .

Then

$\frac{x^{2}}{0.1} = 6.23 \times 10^{-8}$ $x^2/0.1 = 6.23 × 10^"-8"$

$x^{2} = 0.1 \times 6.23 \times 10^{-8} = 6.2 \times 10^{-9}$ $x^2 = 0.1 × 6.23 × 10^"-8" = 6.2 × 10^"-9"$

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

$[H_{3} O^{+}] = x l mol/L = 7.9 \times 10^{-5} l mol/L$ $["H"_3"O"^"+"] = x color(white)(l)"mol/L" = 7.9 × 10^"-5"color(white)(l)"mol/L"$

$pH = -log [H_{3} O^{+}] = -log (7.9 \times 10^{-5}) = 4.10$ $"pH" = "-log"["H"_3"O"^"+"] = "-log"(7.9 × 10^"-5") = 4.10$

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