Calculate the number of moles of hcl(g) that must be added to 1.0 l of 1.0 m nac2h3o2 to produce a solution buffered at each ph.

Missing question: 
(a) pH = pKa.
(b) pH = 3.84.
(c) pH = 5.26.
a) c(sodium acetate - NaC₂H₃O₂) = 1,0 M = 1,0 mol/L.
c(NaC₂H₃O₂) = 1,0 L.
Chemical reaction: CH₃COONa + HCl ⇄ CH₃COOH + NaCl.
Ka - acid dissociation constant.
pKa = -logKa.
When pH = pKa, n(CH₃COO⁻) = n(CH₃COOH).
From chemical reaction, half of sodium acetate reacts (0,5 mol), so n(HCl) = 0,5 mol.

b) Chemical reaction: CH₃COO⁻ + H⁺ ⇄ CH₃COOH.
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∧(-0,54) = 0,288.

(1 mol - x )/x = 0,288.

x = n(HCl) = 0,776 mol.

c) Chemical reaction: CH₃COO⁻ + H⁺ ⇄ CH₃COOH.

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∧(0,50) = 3,16.

(1 mol - x )/x = 3,16.

x = n(HCl) = 0,240 mol.

pH is a numeric scale used to specify the acidity or basicity of an aqueous solution.

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