Question

start: \ce{h2s} 1.000 m + 2 \ce{h2o} \
ightleftharpoons 2 \ce{h3o+} 0 + \ce{s^2-} 0
equil: \underline{0.9923 m} 0.0077 m \quad \quad \quad 0.0154 m \quad 0.0077 m
calculate the $k_a$ from the above data.
the amount of \ce{h2s} that was converted into product is 1 m
the stoichiometry tells us that 1 mol \ce{h2s} will give us 2 mol of \ce{h3o+} and 2 mol of \ce{s^2-}
thus, 3 m \ce{h2s} will give 4 m \ce{h3o+} and 5 m \ce{s^2-}
$k_a = \frac{(__6__)^2 (__7__) }{__9__} = __8__$
the concentration of $\ce{h3o+}$ is squared because its coefficient is 2
a. acetic \quad b. 0.9960 \quad c. 0.0040 \quad d. \ce{al^3+} \quad e. \ce{co2} \quad f. \ce{caf2(aq)}
g. \ce{hf(aq)} \quad h. \ce{caco3(s)} \quad i. \ce{oh-} \quad j. \ce{h3o+} \quad k. \ce{so4^2-} \quad l. \ce{mg^2+}
m. \ce{cl-} \quad n. \ce{hpo4^2-} \quad o. \ce{h2po4-} \quad p. \ce{co3^2-} \quad q. \ce{hco3-} \quad r. \ce{h2s}
s. \ce{hs-} \quad t. \ce{s^2-} \quad u. \ce{h2o(liq)} \quad v. \ce{h2co3} \quad w. \ce{po4^3-} \quad x. \ce{hc2h3o2}
y. \ce{c2h3o2-} \quad z. 1 \times 10^{-14} \quad aa. 0.9923 \quad bb. 0.0077 \quad cc. 3
dd. 2 \quad ee. 1 \quad ff. 0.0154 \quad gg. 3.95 \times 10^{-3} \quad hh. 1.84 \times 10^{-5}

Explanation:

Step1: Find moles of \( H_2S \) converted

Initial \( [H_2S] = 1.000 \, M \), equilibrium \( [H_2S] = 0.9923 \, M \). Converted = \( 1.000 - 0.9923 = 0.0077 \, M \)? Wait, no, wait the start and equil: start \( H_2S = 1.000 \, M \), equil \( H_2S = 0.9923 \, M \), so converted is \( 1.000 - 0.9923 = 0.0077 \, M \)? Wait the first blank (1) – wait maybe I misread. Wait the table: start \( H_2S = 1.000 \, M \), equil \( H_2S = 0.9923 \, M \), and also equil \( HS^- = 0.0077 \, M \)? Wait the reaction: \( H_2S + 2H_2O
ightleftharpoons 2H_3O^+ + S^{2-} \)? Wait no, maybe the correct reaction is \( H_2S
ightleftharpoons H^+ + HS^- \), but the given reaction is \( H_2S + 2H_2O
ightleftharpoons 2H_3O^+ + S^{2-} \). Wait start: \( H_2S = 1.000 \, M \), \( H_3O^+ = 0 \), \( S^{2-} = 0 \). Equil: \( H_2S = 0.9923 \, M \), \( H_3O^+ = 0.0154 \, M \), \( S^{2-} = 0.0077 \, M \), and \( HS^- = 0.0077 \, M \)? Wait the first question: "The amount of \( H_2S \) that was converted into product is 1 M". Converted \( H_2S \) is initial - equil = \( 1.000 - 0.9923 = 0.0077 \, M \)? So 1: BB (0.0077).

Step2: Stoichiometry of \( S^{2-} \)

Reaction: \( H_2S + 2H_2O
ightleftharpoons 2H_3O^+ + S^{2-} \). So 1 mol \( H_2S \) gives 2 mol \( H_3O^+ \) and 1 mol \( S^{2-} \). So blank 2: EE (1).

Step3: Moles from stoichiometry

From step1, converted \( H_2S = 0.0077 \, M \) (BB). So 3: BB (0.0077). Then, 2 mol \( H_3O^+ \) per 1 mol \( H_2S \), so \( [H_3O^+] = 2 \times 0.0077 = 0.0154 \, M \)? Wait equil \( [H_3O^+] \) is 0.0154 M (FF). So 4: FF (0.0154). \( [S^{2-}] = 0.0077 \, M \) (BB), so 5: BB (0.0077).

Step4: \( K_a \) expression

\( K_a = \frac{[H_3O^+]^2 [S^{2-}]}{[H_2S]} \). So 6: FF (0.0154), 7: BB (0.0077), 9: AA (0.9923).

Step5: Calculate \( K_a \)

\( [H_3O^+] = 0.0154 \, M \), \( [S^{2-}] = 0.0077 \, M \), \( [H_2S] = 0.9923 \, M \).

\( K_a = \frac{(0.0154)^2 \times 0.0077}{0.9923} \)

Calculate numerator: \( (0.0154)^2 = 0.00023716 \), times 0.0077: \( 0.00023716 \times 0.0077 \approx 1.826 \times 10^{-6} \)

Divide by 0.9923: \( \approx \frac{1.826 \times 10^{-6}}{0.9923} \approx 1.84 \times 10^{-6} \)? Wait no, wait 0.0154 squared is (0.0154)(0.0154)=0.00023716. Multiply by 0.0077: 0.00023716*0.0077=0.000001826. Divide by 0.9923: ~0.00000184, which is \( 1.84 \times 10^{-6} \)? Wait but let's check the options. HH is \( 1.84 \times 10^{-6} \), so 8: HH.

Wait let's recheck:

1. Converted \( H_2S \): 1.000 - 0.9923 = 0.0077 (BB) – correct.

1. Stoichiometry \( S^{2-} \): 1 mol (EE) – correct.

1. Converted \( H_2S \) is 0.0077 (BB) – correct.

1. \( [H_3O^+] = 0.0154 \) (FF) – correct (since 2*0.0077=0.0154).

1. \( [S^{2-}] = 0.0077 \) (BB) – correct.

1. \( [H_3O^+] = 0.0154 \) (FF) – correct.

1. \( [S^{2-}] = 0.0077 \) (BB) – correct.

1. \( [H_2S] = 0.9923 \) (AA) – correct.

\( K_a = \frac{(0.0154)^2 (0.0077)}{0.9923} \)

Calculate:

\( (0.0154)^2 = 0.00023716 \)

\( 0.00023716 \times 0.0077 = 0.000001826 \)

\( 0.000001826 / 0.9923 ≈ 1.84 \times 10^{-6} \) (HH) – correct.

Answer:

1. BB. 0.0077
2. EE. 1
3. BB. 0.0077
4. FF. 0.0154
5. BB. 0.0077
6. FF. 0.0154
7. BB. 0.0077
8. HH. \( 1.84 \times 10^{-6} \)
9. AA. 0.9923