Answer to Question #88869 in Inorganic Chemistry for Jackson

Question

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

Answer 1

$$NaOH +HCl\rightarrow NaCl + H_2O$$

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:

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