Question

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

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

Homework Answers

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

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