Question

A 25.0mL of 0.152M H2SO4 was reacted with 20.0mL of 0.410 M NaOH.
Calculate the concentration of the salt that is formed during the reaction as well as the concentrations of H2SO4 and NaOH that remain in solution.

Answer 1

Solution :-

Balanced reaction equation

H2SO4 + 2 NaOH ------ > Na2SO4 + 2H2O

now lets calculate the moles of the H2SO4 and NaOH using their molarities and volumes

moles = molarity * volume in liter

moles of H2SO4 = 0.152 mol per L * 0.025 L = 0.0038 mol

moles of NaOH = 0.410 mol per L * 0.020 L = 0.0082 mol

now lets find the moles of NaOH needed to react with the 0.0038 mol H2SO4

0.0038 mol H2SO4 * 2 mol NaOH / 1 mol H2SO4 = 0.0076 mol NaOH

therefore moles of NaOH are excess

lets find the excess moles of NaOH remaining after the reaction

moles of NaOH remain = 0.0082 mol - 0.0076 mol = 0.0006 mol NaOH

now lets calculate the molarity of the NaOH remain after the reaction

molarity of NaOH = moles / total volume

                         = 0.0006 mol / (0.025 L +0.02L)

                         = 0.0133 M

H2SO4 is limiting reactant so it get used up completely so H2SO4 concnetration after the reaction is 0 M

now lets calculat the moles of the salt Na2SO4

mole ratio of the H2SO4 and Na2SO4 is 1 : 1

therefore

0.0038 mol H2SO4 * 1 mol Na2SO4 / 1 mol H2SO4 = 0.0038 mol Na2SO4

so molarity of the salt after th ereaction is [Na2SO4] = 0.0038 mol / (0.025 L + 0.02L)

                                                                            =0.0844 M

Therefore after the reaction concnetrations are as follows

[H2SO4 ] = 0 M

[NaOH] = 0.0133 M

[Na2SO4] =0.0844 M