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Titrasi Asam Basa Contoh Soal -

pOH = (-\log(5.27 \times 10^-6) = 5.28) pH = (14 - 5.28 = 8.72)

For A⁻ + H₂O ⇌ HA + OH⁻, Kb = Kw/Ka = (1.0 \times 10^-14 / 1.8 \times 10^-5 = 5.56 \times 10^-10)

Reaction: ( H_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O ) Here, ( n_a = 2 ) (H₂SO₄ gives 2 H⁺), ( n_b = 1 ) (KOH gives 1 OH⁻). titrasi asam basa contoh soal

0.125 M NaOH Problem 2: Finding Volume (Medium) Question: How many mL of 0.250 M H₂SO₄ are needed to neutralize 50.0 mL of 0.100 M KOH?

[ M_a V_a \times n_a = M_b V_b \times n_b ] [ (0.250)(V_a)(2) = (0.100)(50.0)(1) ] [ 0.500 , V_a = 5.00 ] [ V_a = 10.0 , mL ] pOH = (-\log(5

At equivalence, moles of acid = moles of base = (0.0500 , L \times 0.100 , M = 0.00500 , mol) Total volume = (50.0 + 50.0 = 100.0 , mL = 0.100 , L) Concentration of conjugate base (A⁻) = (0.00500 / 0.100 = 0.0500 , M)

[OH⁻] = (\sqrtK_b \times C = \sqrt(5.56 \times 10^-10)(0.0500)) = (\sqrt2.78 \times 10^-11 = 5.27 \times 10^-6 , M) mL ] At equivalence

[ M_acid V_acid = M_base V_base ] [ (0.100 , M)(25.0 , mL) = M_base (20.0 , mL) ] [ M_base = \frac2.5020.0 = 0.125 , M ]

Reaction: ( HCl + NaOH \rightarrow NaCl + H_2O ) (1:1 ratio)