Question

What volume of 0.250m nitric acid is required to neutralize 4.8mL of 0.150m sodium carbonate?

Answer 1

we have the Balanced chemical equation as:

Na₂CO₃ + 2HNO₃ → 2NaNO₃ + CO₂ + H₂O

Here:

M(Na2CO3)=0.15 M

M(HNO3)=0.25 M

V(Na2CO3)=4.8 mL

According to balanced reaction:

2*number of mol of Na2CO3 =1*number of mol of HNO3

2*M(Na2CO3)*V(Na2CO3) =1*M(HNO3)*V(HNO3)

2*0.15 M *4.8 mL = 1*0.25M *V(HNO3)

V(HNO3) = 5.76 mL

Answer:5.76 mL