Question

Suppose you titrate 80.0 mL of 2.00 M NaOH with 20.0 mL of 4.00 M HCl. What is the final concentration of OH− ions.

Answer 1

Total volume = 80 mL + 20 mL = 100 mL = 0.10 L. 

NaOH + HCl ----> NaCl + H2O. 

0.08 L * 2.00 mol NaOH/L = 0.16 moles NaOH. 

0.02 L * 4.00 mol HCl/L = 0.08 moles HCl. 

0.16 - 0.08 = 0.08. There are 0.08 moles NaOH XS. 

0.08 moles NaOH produces 0.08 moles OH-. 

Concentration OH- = 0.08 moles / 0.10 L = 0.8 M.

Answer 2

Answer: The final concentration of $OH^−$ is 0.8 M.

Explanation: $NaOH\rightarrow Na^++OH^-$

${\text{Moles of NaOH}}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of NaOH}}=2M\times 0.08L=0.16moles$

0.06 moles of NaOH will give 0.06 moles of $[OH^-]$

$HCl\rightarrow H^++Cl^-$

${\text{Moles of HCl}}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of HCl}}=4.0M\times 0.02L=0.08moles$

0.08 moles of HCl will give 0.08 moles of $[H^+]$

$HCl+NaOH\rightarrow NaCl+H_2O$

0.08 moles of $H^+$ will react with 0.08 moles of $OH^-$ and (0.16-0.08)= 0.08 moles of $OH^-$ will be left in 100 ml of solution.

Thus Molarity of $[OH^]-=\frac{moles}{\text {Volume in L}}=\frac{0.08}{0.1}=0.8M$