Question

50.0 mL of 0.200 M NaOH neutralized 20.0 mL of sulfuric acid. Determine the concentration of the acid.

Answer 1

QUESTION # 136686

50.0 mL of 0.200 M NaOH neutralized 20.0 mL of sulfuric acid. Determine the concentration of the acid.

ANSWER

Moles of NAOH used in the reaction is;

0.200 moles are present in 1000mL

X moles are present in 50mL

X=(0.200molx50 mL)/1000mL=0.010moles

Equation for the reaction is

H₂SO₄+2NaOH ⟶Na₂SO₄+2H₂O

From the equation mole ratio of NaOH : H₂SO₄

2 : 1

0.010 : X

Moles of sulphuric acid in reaction is X

X=( 0.010x1)/2=0.005moles

Concentration of the acid is;

0.005moles are present in 20mL

X moles are present in 1000mL

X=(0.005molx1000mL)/20mL= 0.250 moles/L

=0.250M