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.
The number of moles of HCl (g) added at pH 4.20 = 0.78 mole
Given data :
volume = 1 .0 L
Pka of CH₃COOH = 4.75
1.0 L of 1.0m NaC₂H₃O₂ contains 1 mole of CH₃COONa
At pH = 4.20
Given that :
pH = pKa + log [tex]\frac{CH_{3}COONa }{CH_{3}COOH }[/tex]
Above relation can be rewritten as
4.20 = 4.75 + log (1 - x ) / x
therefore
log (1 - x ) / x = -0.55
(1 - x ) / x = Antilog -0.55
x = 0.78
Hence the number of moles of HCL that must be added at pH 4.20 is 0.78 mole
Learn more about Buffer solution : https://brainly.com/question/22390063
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