Answer to Question #212152 in General Chemistry for 22013888

Question

25.0cm3 of 0.200mol dm-3 sodium hydroxide reacted with 28.7 cm3 of sulphuric acid, calculate the concentration of the sulphuric acid. with calculations

Thank you

Answer 1

The equation for the reaction involved is;

$2NaOH+H_2SO_4 \to Na_2SO_4+2H_2O$

Number of moles of NaOH = (25.0/1000)x0.100 = 0.00250 moles of NaOH

find the number of moles of H₂SO₄ are required from to neutralize this NaOH

mole ratio $NaOH:H_2SO_4$ is 2:1

therefore number of moles of H₂SO₄ = $0.00250/2=$ 0.00125 moles H₂SO₄

concentration of $H_2SO_4$ $=$ 0.00125/(28.7/1000) = 0.0436 moldm⁻³