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The composition of an acidic buffer mixt...

The composition of an acidic buffer mixture made up of HA and NaA of total molarity `0.29` having `pH=4.4` and `K_(a)=1.8xx10^(-5)` in terms of concentration of salt and acid respectively is:

A

`0.09M and 0.20M`

B

`0.20M and 0.09M`

C

`0.1M and 0.19M`

D

`0.19M and 0.10M`

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To solve the problem of finding the composition of an acidic buffer mixture made up of HA (acid) and NaA (salt) with a total molarity of 0.29, pH of 4.4, and \( K_a = 1.8 \times 10^{-5} \), we can follow these steps: ### Step 1: Calculate \( pK_a \) The first step is to calculate \( pK_a \) from the given \( K_a \): \[ pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) \] Calculating this gives: \[ pK_a \approx 4.74 \] ### Step 2: Use the Henderson-Hasselbalch Equation The Henderson-Hasselbalch equation for an acidic buffer is given by: \[ pH = pK_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \] Substituting the known values: \[ 4.4 = 4.74 + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \] ### Step 3: Rearranging the Equation Rearranging the equation to isolate the log term: \[ \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = 4.4 - 4.74 = -0.34 \] ### Step 4: Exponentiating to Remove the Logarithm Exponentiating both sides to eliminate the logarithm: \[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{-0.34} \approx 0.46 \] ### Step 5: Set Up the Concentration Relationships Let the concentration of the salt (NaA) be \( [\text{A}^-] = A \) and the concentration of the acid (HA) be \( [\text{HA}] = 0.29 - A \). Thus, we can write: \[ \frac{A}{0.29 - A} = 0.46 \] ### Step 6: Solve for A Cross-multiplying gives: \[ A = 0.46(0.29 - A) \] Expanding this: \[ A = 0.1334 - 0.46A \] Combining like terms: \[ A + 0.46A = 0.1334 \] \[ 1.46A = 0.1334 \] \[ A \approx \frac{0.1334}{1.46} \approx 0.0913 \] ### Step 7: Calculate Concentration of Acid Now, substituting back to find the concentration of the acid: \[ [\text{HA}] = 0.29 - A \approx 0.29 - 0.0913 \approx 0.1987 \] ### Final Answer Thus, the concentrations of the salt and acid in the buffer mixture are approximately: - Concentration of salt (NaA): \( 0.0913 \, \text{M} \) - Concentration of acid (HA): \( 0.1987 \, \text{M} \) ### Summary The composition of the acidic buffer mixture is: - Salt (NaA): \( 0.0913 \, \text{M} \) - Acid (HA): \( 0.1987 \, \text{M} \)

To solve the problem of finding the composition of an acidic buffer mixture made up of HA (acid) and NaA (salt) with a total molarity of 0.29, pH of 4.4, and \( K_a = 1.8 \times 10^{-5} \), we can follow these steps: ### Step 1: Calculate \( pK_a \) The first step is to calculate \( pK_a \) from the given \( K_a \): \[ pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) \] Calculating this gives: ...
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