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_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 ] titrasi asam basa contoh soal
[OH⁻] = (\sqrtK_b \times C = \sqrt(5.56 \times 10^-10)(0.0500)) = (\sqrt2.78 \times 10^-11 = 5.27 \times 10^-6 , M) What is the molar mass of the acid
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? mL) = M_base (20.0
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⁻).