Answer to Question #88869 in Inorganic Chemistry for Jackson

Question #88869
When 25.00mL of sodium hydroxide solution of unknown concentration was titrated against a standard hydrochloric acid solution of concentration 0.098 mol L-1, 14.73 mL of the acid was needed to completely neutralize the base. Calculate the concentration of the sodium hydroxide solution
1
Expert's answer
2019-05-02T07:30:35-0400
"NaOH +HCl\\rightarrow NaCl + H_2O"

Find moles of HCl:


"c= \\frac{n}{V}"

"\\therefore n= c\\times V"

"n(HCl)= 0.098\\frac{mol}{L}\\times14.73\\times10^{-3}L=1.444\\times10^{-3} mol"

According to equation "n(HCl):n(NaOH)=1:1" , then "n(NaOH)=n(HCl)=1.444\\times10^{-3} mol"

Find concentration of NaOH solution:


"c= \\frac{n}{V}= \\frac{1.444\\times10^{-3}mol}{25.00\\times 10^{-3}L}=0.058\\frac{mol}{L}"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS