Answer
The pH of this buffer is equal to 8.95;
Work Step by Step
1000ml = 1L
250ml = 0.25 L
125ml = 0.125 L
1. Find the numbers of moles:
$C(N{H_4}^+) * V(N{H_4}^+) = 0.1*0.25 = 0.025$ moles
$C(NH_3) * V(NH_3) = 0.1*0.125 = 0.0125$ moles
2. Calculate the total volume:
- Total volume: 0.25 + 0.125 = 0.375L
3. Calculate the final concentrations:
$[N{H_4}^+] : \frac{0.025}{0.375} = 0.0667M $
$[N{H_3}] : \frac{0.0125}{0.375} = 0.0333M $
4. Since $N{H_4}^+$ is the conjugate acid of $NH_3$ , we can calculate its kb by using this equation:
$K_b * K_a = K_w = 10^{-14}$
$ 1.76\times 10^{- 5} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 1.76\times 10^{- 5}}$
$K_a = 5.68\times 10^{- 10}$
5. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 5.68 \times 10^{- 10})$
$pKa = 9.25$
6. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.0333}{0.0667}$
- 0.5: It is.
7. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.0333}{5.68 \times 10^{-10}} = 5.86\times 10^{7}$
- $ \frac{0.0667}{5.68 \times 10^{-10}} = 1.17\times 10^{8}$
8. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 9.25 + log(\frac{0.0333}{0.0667})$
$pH = 9.25 + log(0.5)$
$pH = 9.25 + -0.301$
$pH = 8.95$
