Missing question: <span>
(a) pH = pKa.
(b) pH = 3.84.
(c) pH = 5.26.
a) c(sodium acetate - NaC</span>₂H₃O₂) =
1,0 M = 1,0 mol/L.<span>
c(NaC</span>₂H₃O₂) = 1,0 L.<span>
Chemical reaction: CH</span>₃COONa + HCl ⇄ CH₃COOH
+ NaCl.<span>
Ka - acid dissociation constant.
pKa = -logKa.
When pH = pKa, n(CH</span>₃COO⁻) =
n(CH₃COOH).<span>
From chemical reaction, half of sodium
acetate reacts (0,5 mol), so n(HCl) = 0,5 mol.</span>
b) Chemical reaction: CH₃COO⁻ + H⁺ ⇄ CH₃COOH.<span>
Henderson–Hasselbalch equation: pH = pKa
+ log(cs/ck).
4,20 = 4,76 + log(cs/ck).
log(cs/ck) =4,20 - 4,76 = -0,54.
cs/ck = 10</span>∧(-0,54) = 0,288.
(1 mol - x )/x = 0,288.
x = n(HCl) = 0,776 mol.
c) Chemical reaction:
CH₃COO⁻ + H⁺ ⇄ CH₃COOH.
<span>
Henderson–Hasselbalch equation: pH = pKa + log(cs/ck).
5,26 = 4,76 + log(cs/ck).
log(cs/ck) = 5,26 - 4,76 = 0,50.
cs/ck = 10</span>∧(0,50) = 3,16.
(1 mol - x )/x =
3,16.
x = n(HCl) =
0,240 mol.
pH<span> </span><span>is a numeric scale used to specify the </span>acidity<span> or </span>basicity<span> of an </span><span>aqueous solution.</span>