A 14.50 mL sample of nitric acid (HNO3) is titrated to the end point by the addition of 10.45 mL of a 1.525 M solution of barium hydroxide (Ba(OH)2). What is the molarity of the nitric acid solution? (Balanced equation: 2HNO3 + Ba(OH)2 = Ba(NO3)2 + 2 H20)
Balanced chemical equation is:
Ba(OH)2 + 2 HNO3 ---> Ba(NO3)2 + 2 H2O
Here:
M(Ba(OH)2)=1.525 M
V(Ba(OH)2)=10.45 mL
V(HNO3)=14.5 mL
According to balanced reaction:
2*number of mol of Ba(OH)2 =1*number of mol of HNO3
2*M(Ba(OH)2)*V(Ba(OH)2) =1*M(HNO3)*V(HNO3)
2*1.525*10.45 = 1*M(HNO3)*14.5
M(HNO3) = 2.1981 M
Answer: 2.198 M
A 14.50 mL sample of nitric acid (HNO3) is titrated to the end point by the...
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