Question

calculate the ph of the buffered solution if at equilibrium h₂so₃ = 3.50 × 10⁻⁴ m, hso₃⁻¹ = 5.5 × 10⁻² m, and kₐ = 1.75 × 10⁻⁶?
everything is based upon the acid dissolving in water:
h₂so₃ + h₂o <-> hso₃⁻¹ + h₃o⁺¹
first place symbols in the expression for kₐ. use the order given in the reaction.
hso₃⁻¹ 1
kₐ = -------------
2
second, solve for h₃o⁺¹ using the symbols, then put in the numbers.
kₐ 3 (4) (5)
h₃o⁺¹ = ----------- = ------------- = 6
hso₃⁻¹ 7
ph = - log 8 = 9
a. h₃o⁺¹ b. oh⁻¹ c. h₂c₆h₆o₆ d. hc₆h₆o₆⁻¹ e. hso₃⁻¹ f. so₃⁻²
g. h₂so₃ h. hc₂h₃o₂ i. c₂h₃o₂⁻¹ j. h₂s k. hs⁻¹ l. s⁻²
m. hx n. x⁻¹ o. 1 p. 2 q. 3 r. 0.5000 s. 5.50 × 10⁻⁵
t. 3.50 × 10⁻⁴ u. 8.00 × 10⁻⁵ v. 1.75 × 10⁻⁶ w. 1.11 × 10⁻⁸ x. 5.50 × 10⁻²

Explanation:

Step1: Write \( K_a \) expression

For the reaction \( \ce{H2SO3 + H2O <-> HSO3^- + H3O^+} \), the acid dissociation constant \( K_a \) is \( K_a=\frac{[\ce{HSO3^-}][\ce{H3O^+}]}{[\ce{H2SO3}]} \). So blank 1 is \( \ce{H3O^+} \) (option A) and blank 2 is \( \ce{H2SO3} \) (option G).

Step2: Solve for \( [\ce{H3O^+}] \)

Rearrange \( K_a \) formula: \( [\ce{H3O^+}]=\frac{K_a[\ce{H2SO3}]}{[\ce{HSO3^-}]} \). So blank 3 is \( [\ce{H2SO3}] \) (option G), blank 4 is \( K_a = 1.75\times10^{-6} \) (option V), blank 5 is \( [\ce{H2SO3}]=3.50\times10^{-4} \) (option T), blank 7 is \( [\ce{HSO3^-}]=5.5\times10^{-2} \) (option X). Calculate numerator: \( (1.75\times10^{-6})(3.50\times10^{-4}) = 6.125\times10^{-10} \), denominator: \( 5.5\times10^{-2} \), so \( [\ce{H3O^+}]=\frac{6.125\times10^{-10}}{5.5\times10^{-2}}\approx1.11\times10^{-8} \)? Wait, no, wait: \( K_a = 1.75\times10^{-6} \), \( [\ce{H2SO3}]=3.50\times10^{-4} \), \( [\ce{HSO3^-}]=5.5\times10^{-2} \). So \( [\ce{H3O^+}]=\frac{1.75\times10^{-6}\times3.50\times10^{-4}}{5.5\times10^{-2}}=\frac{6.125\times10^{-10}}{5.5\times10^{-2}}\approx1.11\times10^{-8} \)? Wait, no, miscalculation: \( 1.75\times3.50 = 6.125 \), \( 10^{-6}\times10^{-4}=10^{-10} \), denominator \( 5.5\times10^{-2} \), so \( 6.125\times10^{-10}/5.5\times10^{-2}= (6.125/5.5)\times10^{-8}\approx1.11\times10^{-8} \)? Wait, no, 6.125/5.5≈1.11, so \( [\ce{H3O^+}]\approx1.11\times10^{-8} \)? Wait, no, wait: \( K_a = 1.75\times10^{-6} \), \( [\ce{H2SO3}]=3.50\times10^{-4} \), \( [\ce{HSO3^-}]=5.5\times10^{-2} \). So \( [\ce{H3O^+}]=\frac{K_a\times[\ce{H2SO3}]}{[\ce{HSO3^-}]}=\frac{1.75\times10^{-6}\times3.50\times10^{-4}}{5.5\times10^{-2}} \). Calculate: \( 1.75\times3.50 = 6.125 \), \( 10^{-6}\times10^{-4}=10^{-10} \), \( 6.125\times10^{-10}/5.5\times10^{-2}= (6.125/5.5)\times10^{-8}\approx1.11\times10^{-8} \)? Wait, but let's recalculate: 6.125 ÷ 5.5 = 1.1136..., so \( [\ce{H3O^+}]\approx1.11\times10^{-8} \) (option W). Wait, but maybe I made a mistake. Wait, \( K_a = 1.75\times10^{-6} \), \( [\ce{H2SO3}]=3.50\times10^{-4} \), \( [\ce{HSO3^-}]=5.5\times10^{-2} \). So \( [\ce{H3O^+}]=\frac{1.75\times10^{-6}\times3.50\times10^{-4}}{5.5\times10^{-2}}=\frac{6.125\times10^{-10}}{5.5\times10^{-2}} = 1.1136\times10^{-8} \approx1.11\times10^{-8} \) (option W). So blank 6 is \( 1.11\times10^{-8} \) (option W).

Step3: Calculate pH

\( pH = -\log[\ce{H3O^+}] \), so blank 8 is \( [\ce{H3O^+}]=1.11\times10^{-8} \) (option W), and \( pH = -\log(1.11\times10^{-8})\approx7.95 \) (but let's check the calculation again. Wait, maybe I messed up the formula. Wait, the buffer is \( \ce{H2SO3/HSO3^-} \), but \( K_a = 1.75\times10^{-6} \), \( [\ce{H2SO3}]=3.50\times10^{-4} \), \( [\ce{HSO3^-}]=5.5\times10^{-2} \). Wait, \( [\ce{H2SO3}] \) is smaller than \( [\ce{HSO3^-}] \), so \( [\ce{H3O^+}]=K_a\times\frac{[\ce{acid}]}{[\ce{conjugate base}]} = 1.75\times10^{-6}\times\frac{3.50\times10^{-4}}{5.5\times10^{-2}} \). Let's compute \( \frac{3.50\times10^{-4}}{5.5\times10^{-2}}=\frac{3.50}{5.5}\times10^{-2}\approx0.636\times10^{-2}=6.36\times10^{-3} \). Then \( 1.75\times10^{-6}\times6.36\times10^{-3}=1.11\times10^{-8} \). Then \( pH = -\log(1.11\times10^{-8})\approx7.95 \). But let's check the steps again.

Now, let's fill the blanks:

1. A (H₃O⁺)
2. G (H₂SO₃)
3. G ([H₂SO₃])
4. V (1.75×10⁻⁶)
5. T (3.50×10⁻⁴)
6. W (1.11×10⁻⁸)
7. X (5.5×10⁻²)
8. W ([H₃O⁺] = 1.11×10⁻⁸)
9. Let's calculate \( -\log(1.11\times10^{-8}) \). \( \log(1.11\times10^{-8})=\log(1.11) + \log(10^{-8})\approx0.0453 - 8=-7.9547 \), so \( pH\approx7.95 \) (but maybe the options have a different value? Wait, maybe I made a mistake in the formula. Wait, the reaction is \( \ce{H2SO3 <-> HSO3^- + H^+} \) (or \( \ce{H3O^+} \)), so \( K_a=\frac{[\ce{HSO3^-}][\ce{H^+}]}{[\ce{H2SO3}]} \), so \( [\ce{H^+}]=K_a\times\frac{[\ce{H2SO3}]}{[\ce{HSO3^-}]} \). So \( [\ce{H2SO3}]=3.50\times10^{-4} \), \( [\ce{HSO3^-}]=5.5\times10^{-2} \), \( K_a = 1.75\times10^{-6} \). So \( [\ce{H^+}]=1.75\times10^{-6}\times\frac{3.50\times10^{-4}}{5.5\times10^{-2}} = 1.75\times3.50\times10^{-10}/5.5\times10^{-2}=6.125\times10^{-10}/5.5\times10^{-2}=1.11\times10^{-8} \). Then \( pH = -\log(1.11\times10^{-8})\approx7.95 \). But let's check if there's a miscalculation. Wait, 3.50e-4 divided by 5.5e-2 is (3.50/5.5)e-2 ≈ 0.636e-2 = 6.36e-3. Then 1.75e-6 * 6.36e-3 = 1.11e-8. Then log(1.11e-8) = log(1.11) + log(1e-8) ≈ 0.0453 - 8 = -7.9547, so pH ≈ 7.95.

Now, let's list the answers for each blank:

1. A
2. G
3. G
4. V
5. T
6. W
7. X
8. W
9. ~7.95 (but let's check the options. Wait, maybe I made a mistake in the formula. Wait, the buffer is a weak acid - conjugate base pair. The Henderson - Hasselbalch equation is \( pH = pK_a+\log\frac{[\ce{base}]}{[\ce{acid}]} \). Let's use that. \( pK_a = -\log(1.75\times10^{-6})\approx5.757 \). \( \log\frac{5.5\times10^{-2}}{3.50\times10^{-4}}=\log(\frac{5.5}{3.50}\times10^{2})=\log(1.571\times10^{2})=\log(1.571)+2\approx0.196 + 2 = 2.196 \). Then \( pH = 5.757+2.196\approx7.953 \), which matches the previous calculation. So pH is approximately 7.95.

Answer:

1. A. \( \ce{H3O^+1} \)
2. G. \( \ce{H2SO3} \)
3. G. \( \ce{H2SO3} \)
4. V. \( 1.75\times10^{-6} \)
5. T. \( 3.50\times10^{-4} \)
6. W. \( 1.11\times10^{-8} \)
7. X. \( 5.5\times10^{-2} \)
8. W. \( 1.11\times10^{-8} \)
9. Approximately \( 7.95 \) (if we consider the calculation, but let's check the options. Wait, maybe the question expects the steps as above.