Question

200.0 mL of 0.200 M HCl is titrated with 0.050 M NaOH. What is the pH after the addition of 100. mL of the NaOH solution

Answer 1

Answer : The pH of the solution is, 0.932

Explanation :

First we have to calculate the moles of HCl and NaOH.

`\text{Moles of HCl}=\text{Concentration of HCl}* \text{Volume of solution}=0.200mole/L* 0.200L=0.040mole`

`\text{Moles of NaOH}=\text{Concentration of NaOH}* \text{Volume of solution}=0.050mole/L* 0.100L=0.0050mole`

The balanced chemical reaction will be,

`HCl+NaOH\rightarrow NaCl+H_2O`

From the balanced reaction we conclude that,

As, 1 mole of NaOH neutralizes by 1 mole of HCl

So, 0.0050 mole of NaOH neutralizes by 0.0050 mole of HCl

Thus, the number of neutralized moles = 0.0050 mole

Remaining moles of HCl = 0.040 - 0.0050 = 0.035 moles

Total volume of solution = 200.0 mL + 100.0 mL = 300.0 mL = 0.300 L

Now we have to calculate the concentration of HCl(acid).

`Concentration=(Moles)/(Volume)=(0.035mol)/(0.300L)=0.117M`

As we know that, 1 mole of HCl dissociates to give 1 mole of hydrogen ion and 1 mole of chloride ion.

So, concentration of
`H^+` = 0.117 M

Now we have to calculate the pH of solution.

`pH=-\log [H^+]`

`pH=-\log (0.117)`

`pH=0.932`

Thus, the pH of the solution is, 0.932