Answer to Question #345703 in General Chemistry for Kaushi

Question

What is the total number of moles of ions present in a solution after mixing 125cm3 of 0.2M NaOH and 125cm3 of 1M H2SO4

Answer 1

n = C_M * V in the formula,

n — amount of substance (mol);

C_M — concentration of molarity (mol/l);

V — volume of solution (litr)

1.

n (NaOH) = 0,2 * 0,125 = 0,025 mol NaOH

NaOH => Na⁺ + OH^–

1mol —> 2mol ion

0,025mol —> X

X = 0,025 * 2 / 1 = 0,05mol ion

2.

n (H₂SO₄) = 1 * 0,125 = 0,125 mol H₂SO₄

H₂SO₄ => 2 H⁺ + (SO₄)^2–

1mol —> 3mol ion

0,125mol —> X

X = 0,125 * 3 / 1 = 0,375 mol ion

Total ion = 0,05 + 0,375 = 0,425mol ion

ANSWER: 0,425 mol ion