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
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