Question

A solution of 0.262 M KOH is used to titrate 10.0 mL of a 0.190 M H3PO4 solution. What volume, in milliliters, of the KOH solution is required? H3PO4(aq)+3KOH(aq)→3H2O(l)+K3PO4(aq)

Answer 1

To titrate 10.0 mL of a 0.190 M H3PO4 solution, 72 milliliters of a 0.262 M KOH solution is required.

Step-by-step explanation:

To find the volume of KOH solution required to titrate the H3PO4 solution, we need to use the balanced chemical equation:

H3PO4(aq) + 3KOH(aq) → 3H2O(l) + K3PO4(aq)

The mole ratio between H3PO4 and KOH is 1:3. In this case, the initial concentration of H3PO4 is 0.19 M and the volume used is 10.0 mL (or 0.01 L). Using the equation M1V1 = M2V2, we can calculate the volume of KOH solution required:

M1V1 = M2V2

(0.19 M)(0.01 L) = (0.262 M)(V2)

V2 = (0.19 M)(0.01 L) / (0.262 M)

V2 ≈ 0.072 L

Converting to milliliters, V2 = 0.072 L * 1000 mL/L = 72 mL

Therefore, 72 milliliters of the KOH solution is required to titrate the H3PO4 solution.