Answer
$pH \approx 4.37$
Work Step by Step
1. Calculate the molar mass:
1.01* 1 + 12.01* 7 + 1.01* 5 + 16* 2 = 122.13g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 21.5}{ 122.13}$
$n(moles) = 0.176$
3. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 0.176}{ 0.2} $
$C(mol/L) = 0.8802$
4. Calculate the molar mass:
22.99* 1 + 12.01* 7 + 1.01* 5 + 16* 2 = 144.11g/mol
5. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 37.7}{ 144.11}$
$n(moles) = 0.2616$
6. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 0.2616}{ 0.2} $
$C(mol/L) = 1.308$
7. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 6.4 \times 10^{- 5})$
$pKa = 4.194$
8. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{1.308}{0.8802}$
- 1.486: It is.
9. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{1.308}{6.4 \times 10^{-5}} = 2.044\times 10^{4}$
- $ \frac{0.8802}{6.4 \times 10^{-5}} = 1.375\times 10^{4}$
10. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 4.194 + log(\frac{1.308}{0.8802})$
$pH = 4.194 + log(1.486)$
$pH = 4.194 + 0.172$
$pH = 4.366 \approx 4.37$
