Question

78 mL of 2.5 M phosphoric acid is neutralized with 500 mL potassium hydroxide. What is the concentration of the base? Include the balanced chemical equation for the reaction

Answer 1

Given:

78 mL of 2.5 M phosphoric acid is neutralized with 500 mL potassium hydroxide.

To find:

Concentration of the base.

Calculation:

Since the acid got completely neutralised by the base, we can say that:

n refers to "n - factor" ; S refers to strength of acid/base ; V refers to volume:

$\sf{(n1) \times (S1) \times (V1) = (n2) \times (S2) \times (V2)}$

$\sf{ = > 3 \times 2.5 \times 78= 1 \times (S2)\times 500}$

$\sf{ = > 500 (S2) = 585}$

$\sf{ = > (S2) = \dfrac{ 585}{500}}$

$\sf{ = > (S2) = 1.17 \: M}$

Hence, concentration of base is 1.17 M.

$\boxed{\rm{\large{H_{3}PO_{4} + 3KOH \rightarrow K_{3}PO_{4} + 3H_{2}O}}}$