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* {\text{Volume of solution in liters}}`

`{\text{Moles of NaOH}}=2M* 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* {\text{Volume of solution in liters}}`

`{\text{Moles of HCl}}=4.0M* 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}}=(0.08)/(0.1)=0.8M`