Question #66887
63.9 ml of 0.560 M sulfuric acid solution was completely neutralized by 28.3 ml of potassium hydroxide solution. What was the concentration of the potassium hydroxide?
Expert's answer
Reaction of neutralization of sulfuric acid by potassium hydroxide:
H_2 SO_4+2KOH=K_2 SO_4+H_2 O
Using the Law of equivalence:
n(H_2 SO_4 )=n(KOH)
c(H_2 SO_4 )∙V(H_2 SO_4 )=c(KOH)∙V(KOH)
c(KOH)=(c(H_2 SO_4 )∙V(H_2 SO_4 ))/V(KOH) =(63.6 ml∙0.560 M)/(28.3 ml)=1.259 M
Answer: concentration of potassium hydroxide is 1.259 M.
H_2 SO_4+2KOH=K_2 SO_4+H_2 O
Using the Law of equivalence:
n(H_2 SO_4 )=n(KOH)
c(H_2 SO_4 )∙V(H_2 SO_4 )=c(KOH)∙V(KOH)
c(KOH)=(c(H_2 SO_4 )∙V(H_2 SO_4 ))/V(KOH) =(63.6 ml∙0.560 M)/(28.3 ml)=1.259 M
Answer: concentration of potassium hydroxide is 1.259 M.
