[ 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 ]
10.0 mL of H₂SO₄ Problem 3: Finding Molar Mass of Unknown Acid (Advanced) Question: A 0.200 g sample of an unknown monoprotic acid (HA) is titrated with 0.100 M NaOH. The equivalence point is reached after adding 25.0 mL of NaOH. What is the molar mass of the acid?
pOH = (-\log(5.27 \times 10^-6) = 5.28) pH = (14 - 5.28 = 8.72) titrasi asam basa contoh soal
Reaction: ( HCl + NaOH \rightarrow NaCl + H_2O ) (1:1 ratio)
[OH⁻] = (\sqrtK_b \times C = \sqrt(5.56 \times 10^-10)(0.0500)) = (\sqrt2.78 \times 10^-11 = 5.27 \times 10^-6 , M) [ M_acid V_acid = M_base V_base ] [ (0
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⁻).
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) pOH = (-\log(5
For A⁻ + H₂O ⇌ HA + OH⁻, Kb = Kw/Ka = (1.0 \times 10^-14 / 1.8 \times 10^-5 = 5.56 \times 10^-10)
