# Chapter 15 - Acid-Base Equilibria - Exercises - Page 752: 37

$pH \approx 4.37$

#### 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: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 4.194 + log(\frac{1.308}{0.8802})$ $pH = 4.194 + log(1.486)$ $pH = 4.194 + 0.172$ $pH = 4.366 \approx 4.37$

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