Question

What are the final hydrogen ion concentration and pH of a solution obtained by mixing 400mL of 0.2M NaOH with 150mL of 0.1M H3PO4?

Answer 1

Answer : The final hydrogen ion concentration is
`1.58* 10^(-13)M`

Explanation :

The chemical reaction equation will be:

`H_3PO_4+3NaOH\rightarrow Na_3PO_4+3H_2O`

In this reaction, 1 mole of
`H_3PO_4` reacts with 3 mole NaOH.

So, the number of moles of
`H_3PO_4` present in 150 ml of 0.1 M solution is calculated as follows.

Number of moles = Concentration × Volume

Number of moles = 0.1 M ×0.150 L = 0.015 mol

As it reacts with 3 moles of NaOH.

Number of moles of NaOH = 3 × 0.015 mol = 0.045 mol

So, moles of NaOH in 400 mL of 0.2 M NaOH is as follows.

Number of moles = 0.2 M × 0.4 L = 0.080 mol

Number of moles remained after the reaction = (0.080 - 0.045) mol = 0.035 mol NaOH in 550 ml (400 ml + 150 ml)

As molarity is the number of moles present in liter of solution. Hence, molarity of NaOH is as follows.

`\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution}}`

`\text{Molarity}=(0.035mol)/(0.550L)=0.0636M`

Now we have to determine the hydroxide ion concentration.

As,
`[OH^-]` = 0.0636 M

`pOH=-\log [OH^-]`

`pOH=-\log 0.0636`

`pOH=1.20`

Now we have to determine the pH.

As, pH + pOH = 14

pH = 14 - pOH

pH = 14 - 1.20

pH = 12.8

Now we have to determine the hydrogen ion concentration.

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

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

`[H^+]=1.58* 10^(-13)M`

Therefore, the final hydrogen ion concentration is
`1.58* 10^(-13)M`