Question

calculate the ph of the above buffer after 10.00 ml of 1.000 m naoh (0.01000 mol oh⁻¹) was added to 1.000 l of the buffer solution. the new concentrations are hc₂h₃o₂ = 0.4900 m and c₂h₃o₂⁻¹ = 0.7100 m. the kₐ = 1.60 x 10⁻⁵. everything is based upon the acid dissolving in water: hc₂h₃o₂ + h₂o <-> c₂h₃o₂⁻¹ + h₃o⁺¹ first write the expression for kₐ using symbols c₂h₃o₂⁻¹1 kₐ = ------------- 2 second, solve for h₃o⁺¹ using symbols, then put in the numbers kₐ 3 (4)(5) h₃o⁺¹ = --------- = ----------- = 6 7 8 ph = - log 9 = 10 a. h₃o⁺¹ b. oh⁻¹ c. h₂c₆h₆o₆ d. hc₆h₆o₆⁻¹ e. hso₃⁻¹ f. so₃²⁻ g. h₂so₃ h. hc₂h₃o₂ i. c₂h₃o₂⁻¹ j. h₂s k. hs⁻¹ l. s²⁻ m. hx n. x⁻¹ o. 1 p. 2 q. 3 r. 0.5000 s. 5.50 x 10⁻⁵

Explanation:

Step1: Recall Ka Expression

For the reaction $\ce{HC2H3O2 + H2O <-> C2H3O2^- + H3O^+}$, the acid dissociation constant $K_a$ is given by $K_a = \frac{[\ce{C2H3O2^-}][\ce{H3O^+}]}{[\ce{HC2H3O2}]}$. So blank 1 is $\ce{H3O^+}$ (option A) and blank 2 is $\ce{HC2H3O2}$ (option H).

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

Rearranging the $K_a$ expression: $[\ce{H3O^+}] = \frac{K_a [\ce{HC2H3O2}]}{[\ce{C2H3O2^-}]}$. So blank 3 is $\ce{HC2H3O2}$ (H), blank 4 is $K_a = 1.60\times10^{-5}$, blank 5 is $[\ce{HC2H3O2}] = 0.4900$ M, blank 7 is $[\ce{C2H3O2^-}] = 0.7100$ M.

Calculating numerator: $(1.60\times10^{-5})(0.4900) = 7.84\times10^{-6}$

Denominator: $0.7100$

So $[\ce{H3O^+}] = \frac{7.84\times10^{-6}}{0.7100} \approx 1.104\times10^{-5}$? Wait, no, wait: Wait, $K_a = 1.60\times10^{-5}$, $[\ce{HC2H3O2}] = 0.4900$, $[\ce{C2H3O2^-}] = 0.7100$.

So $[\ce{H3O^+}] = K_a \times \frac{[\ce{HC2H3O2}]}{[\ce{C2H3O2^-}]} = 1.60\times10^{-5} \times \frac{0.4900}{0.7100}$

Calculate $\frac{0.4900}{0.7100} \approx 0.6901$

Then $1.60\times10^{-5} \times 0.6901 \approx 1.104\times10^{-5}$? Wait, maybe I miscalculated. Wait, $1.60\times10^{-5} \times 0.49 = 7.84\times10^{-6}$; $7.84\times10^{-6} / 0.71 \approx 1.104\times10^{-5}$? Wait, no, $0.49/0.71 \approx 0.6901$, $1.6\times0.6901 \approx 1.104$, so $1.104\times10^{-5}$? Wait, but let's check again.

Wait, $K_a = 1.60\times10^{-5}$, $[\ce{HA}] = 0.49$, $[\ce{A^-}] = 0.71$.

$[\ce{H3O^+}] = K_a \times \frac{[\ce{HA}]}{[\ce{A^-}]} = 1.60\times10^{-5} \times \frac{0.4900}{0.7100} = 1.60\times10^{-5} \times 0.69014 \approx 1.104\times10^{-5}$ M.

Then pH = -log(1.104\times10^{-5}) ≈ 4.957.

But let's fill the blanks:

Blank 6: $[\ce{H3O^+}]$ value, blank 8: denominator (0.7100).

Then pH = -log(blank 9, which is $[\ce{H3O^+}]$) to get blank 10.

But let's go step by step with the options.

Wait, the options for numbers: S is $5.50\times10^{-5}$, but our calculation is ~1.10\times10^{-5}? Wait, maybe I made a mistake. Wait, no, the given $K_a$ is $1.60\times10^{-5}$, $[\ce{HC2H3O2}] = 0.49$, $[\ce{C2H3O2^-}] = 0.71$.

Wait, $[\ce{H3O^+}] = K_a \times \frac{[\ce{HA}]}{[\ce{A^-}]} = 1.60e-5 * (0.49 / 0.71) = 1.60e-5 * 0.6901 = 1.104e-5$. Then pH = -log(1.104e-5) ≈ 4.957.

But maybe the options have a typo or I misread. Wait, maybe the original buffer before addition had different concentrations, but the problem says new concentrations are [HC2H3O2] = 0.4900 M and [C2H3O2^-] = 0.7100 M.

So proceeding:

Blank 1: A (H3O+1)

Blank 2: H (HC2H3O2)

Blank 3: H (HC2H3O2)

Blank 4: 1.60×10⁻⁵

Blank 5: 0.4900

Blank 7: I (C2H3O2⁻¹)

Calculating $[\ce{H3O^+}]$: (1.60×10⁻⁵)(0.4900) / 0.7100 = (7.84×10⁻⁶) / 0.7100 ≈ 1.104×10⁻⁵ M. But the options for blank 6: S is 5.50×10⁻⁵, which is not matching. Wait, maybe the $K_a$ is different? Wait, no, the problem says $K_a = 1.60×10⁻⁵$.

Wait, maybe I messed up the ratio. Wait, buffer formula: pH = pKa + log([A⁻]/[HA]). pKa = -log(1.60×10⁻⁵) ≈ 4.796. Then log(0.71/0.49) = log(1.4489) ≈ 0.161. So pH = 4.796 + 0.161 ≈ 4.957, which matches the -log(1.10×10⁻⁵) ≈ 4.957.

But the options for blank 6: maybe the intended calculation was different? Wait, maybe the original buffer had [HA] = 0.5 and [A⁻] = 0.5, but no, the problem says new concentrations are 0.49 and 0.71.

Alternatively, maybe the question has a typo, but let's proceed with the given numbers.

So blank 6: $1.10\times10^{-5}$ (but not in options? Wait, S is 5.50×10⁻⁵, maybe I made a mistake. Wait, no, let's recalculate:

$K_a = 1.60\times10^{-5}$, $[\ce{HA}] = 0.49$, $[\ce{A^-}] = 0.71$

$[\ce{H3O^+}] = (1.60\times10^{-5})(0.49)/0.71 = (7.84\times10^{-6})/0.71 ≈ 1.104\times10^{-5}$ M.

Then pH = -log(1.104×10⁻⁵) ≈ 4.957, so approximately 4.96.

But let's check the options for blank 9: it's $[\ce{H3O^+}]$, so blank 9 is the value of $[\ce{H3O^+}]$, blank 10 is pH.

But maybe the intended answer is using the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA])

pKa = -log(1.60×10⁻⁵) ≈ 4.796

log(0.71/0.49) = log(1.4489) ≈ 0.161

pH ≈ 4.796 + 0.161 ≈ 4.957 ≈ 4.96.

But let's fill the blanks as per the steps:

1. A (H3O+1)

2. H (HC2H3O2)

3. H (HC2H3O2)

4. 1.60×10⁻⁵

5. 0.4900

6. (1.60×10⁻⁵×0.4900)/0.7100 ≈ 1.10×10⁻⁵ (but no option, maybe I misread the concentrations. Wait, maybe the original [HA] was 0.5 and [A⁻] was 0.5, but no, the problem says 0.49 and 0.71.

Wait, maybe the question has a mistake, but assuming the steps:

After solving, the pH is approximately 4.96.

But let's proceed with the given numbers.

Answer:

1. A. $\ce{H3O^+1}$
2. H. $\ce{HC2H3O2}$
3. H. $\ce{HC2H3O2}$
4. $1.60 \times 10^{-5}$
5. $0.4900$ M
6. $\approx 1.10 \times 10^{-5}$ M (or follow calculation)
7. I. $\ce{C2H3O2^-1}$
8. $0.7100$
9. $[\ce{H3O^+}] \approx 1.10 \times 10^{-5}$ M
10. $\approx 4.96$

(Note: If the intended calculation differs, adjust based on correct arithmetic, but the key is following the $K_a$ expression and buffer pH formula.)