Question

EQUL 499: Studying the pH of Strong Acid, Weak Acid, Salt, and Buffer Solurions name 101 section date Data Sheet 1 , Preparing HCI Solutions and Determining pH concentration of HCI, M measured pH theoretical pH Z.04 1.0x 10- 218 3.54 3,91 1.0 x 10- 1.0 x 10- 1.0x 10 I. Preparing HC2H302 Solutions and Determining pH cancentration of HC H,O, M measured pH theoretical pH 3-82 4.22 1.0 x 10 1.0 x 10-2 5.06 1.0 x 10-3 5.71 1.0x 10 calculated K, of HCH,O2 concentration of HC H,02, M literature K of HC H.о. based on pH data O Ha pniv enaltul 1.0x 10-1 1.0 x 10 1.0x 10-3 1.0 x 10

Answer 1

1.

HCl is a strong acid. It dissociate completely as

$HCl_{(aq)} \rightarrow H^{+}_{(aq)} + Cl^{-} _{(aq)}$

ie. [HCl] = [H⁺]

​​​​​​Formula :- $p^{H} = - log[H^{+}]$

For, $[HCl] = 1.0 \times 10^{-1} = 0.1$

  $p^{H} = - log[0.1] = -(-1) = 1$

For, $[HCl] = 1.0 \times 10^{-2} = 0.01$   

  $p^{H} = - log[0.01] = -(-2) = 2$

For, $[HCl] = 1.0 \times 10^{-3} = 0.001$   

$p^{H} = - log[0.001] = -(-3) = 3$

For, $[HCl] = 1.0 \times 10^{-4} = 0.0001$   

$p^{H} = - log[0.0001] = -(-4) = 4$

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2. HC₂H₃O₂ is a weak acid. It is partially dissociate as

$HC_{2}H_{3}O_{2}_{(aq)} \rightleftharpoons H^{+}_{(aq)} + C_{2}H_{3}O_{2}^{-}_{(aq)}$

Now, p^Ka of HC₂H₃O₂ is 4.76

Formula :- $p^{ka} = log[HA] + (2 \times p^{H})$

i.e   $p^{ka} = log[HC_{2}H_{3}O_{2}] + (2 \times p^{H})$

For, 0.1 M

$4.76 = log[0.1] + (2 \times p^{H})$

$4.76 = -1 + (2 \times p^{H})$

$p^{H} = \frac{4.76 + 1}{2} = 2.88$

For, 0.01 M

$4.76 = log[0.01] + (2 \times p^{H})$

$4.76 = -2 + (2 \times p^{H})$

$p^{H} = \frac{4.76 + 2}{2} = 3.38$

For, 0.001 M

$4.76 = log[0.001] + (2 \times p^{H})$

$4.76 = -3 + (2 \times p^{H})$

$p^{H} = \frac{4.76 + 3}{2} = 3.88$

For, 0.0001 M

$4.76 = log[0.0001] + (2 \times p^{H})$

$4.76 = -4 + (2 \times p^{H})$

$p^{H} = \frac{4.76 + 4}{2} = 4.38$

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Ka from p^H data

Formula :-

$p^{H} = - log \sqrt{K_{a} [HA]}$

ie. $K_{a} = \frac{[Antilog(-p^{H})]^{2}}{[HA]}$

For, pH = 3.82

$K_{a} = \frac{[Antilog(-3.82)]^{2}}{0.1} = \frac{[1.51 \times 10^{-4}]^{2}}{0.1} = 2.28 \times 10^{-7}$

For, pH = 4.22

$K_{a} = \frac{[Antilog(-4.22)]^{2}}{0.01} = \frac{[6.03 \times 10^{-5}]^{2}}{0.01} = 3.64 \times 10^{-7}$

For, pH = 5.06

$K_{a} = \frac{[Antilog(-5.06)]^{2}}{0.001} = \frac{[8.71 \times 10^{-6}]^{2}}{0.001} = 7.59 \times 10^{-8}$

For, pH = 5.71

$K_{a} = \frac{[Antilog(-5.72)]^{2}}{0.0001} = \frac{[1.95 \times 10^{-6}]^{2}}{0.0001} = 3.8 \times 10^{-8}$

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Literature Ka

For, pH = 2.88

$K_{a} = \frac{[Antilog(-2.88)]^{2}}{0.1} = \frac{[1.32 \times 10^{-3}]^{2}}{0.1} = 1.74 \times 10^{-5}$

For, pH = 3.38

$K_{a} = \frac{[Antilog(-3.38)]^{2}}{0.01} = \frac{[4.17 \times 10^{-4}]^{2}}{0.01} = 1.74 \times 10^{-5}$

For, pH = 3.88

$K_{a} = \frac{[Antilog(-3.88)]^{2}}{0.001} = \frac{[1.32 \times 10^{-4}]^{2}}{0.001} = 1.74 \times 10^{-5}$

For, pH = 4.38

$K_{a} = \frac{[Antilog(-4.38)]^{2}}{0.0001} = \frac{[4.17 \times 10^{-5}]^{2}}{0.0001} = 1.74 \times 10^{-5}$