Answer
$pH = 10.20$
Work Step by Step
1000ml = 1L
275ml = 0.275 L
175ml = 0.175 L
1. Find the numbers of moles:
$C(C_2H_5N{H_3}^+) * V(C_2H_5N{H_3}^+) = 0.2*0.275 = 0.055$ moles
$C(C_2H_5N{H_2}) * V(C_2H_5N{H_2}) = 0.1*0.175 = 0.0175$ moles
2. Calculate the total volume:
- Total volume: 0.275 + 0.175 = 0.45L
3. Calculate the final concentrations:
$[C_2H_5N{H_3}^+] : \frac{0.055}{0.45} = 0.122M $
$[C_2H_5N{H_2}] : \frac{0.0175}{0.45} = 0.0389M $
4. Since $C_2H_5N{H_3}^+$ is the conjugate acid of $C_2H_5N{H_2}$ , we can calculate its kb by using this equation:
$K_b * K_a = K_w = 10^{-14}$
$ 5.6\times 10^{- 4} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 5.6\times 10^{- 4}}$
$K_a = 1.79\times 10^{- 11}$
5. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 1.79 \times 10^{- 11})$
$pKa = 10.7$
6. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.0389}{0.122}$
- 0.318: It is.
7. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.0389}{1.79 \times 10^{-11}} = 2.18\times 10^{9}$
- $ \frac{0.122}{1.79 \times 10^{-11}} = 6.84\times 10^{9}$
8. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 10.7 + log(\frac{0.0389}{0.122})$
$pH = 10.7 + log(0.318)$
$pH = 10.7 + -0.50$
$pH = 10.20$
