Calculate the pH of the following solutions obtained by mixing :
(a)equal volumes of two solutions with pH=4 and pH=10
(b)400 mL of solution with pH=3 and 100 mL of solution with pH=4
(c )200 mL of 0.1 M NaOH and 300 mL of 0.2 KOH

Explanation

(a)For a solution having pH=4,
`[H_3O^+]=10^(-4)` M
For a solution having pH=10
`[H_3O^+]=10^(-10)` M
`therefore [OH^-]=10^(-14)/10^(-10)=10^(-4)` M
The two solutions will exactly neutralise each other and therefore , the resulting solution will neutral and its pH will be 7.
(b) For a solution having pH=3 , `[H_3O^+]=10^(-3)` M
Conc. of `H_3O^+` in 400 mL =`10^(-3)/1000xx400`
`=4xx10^(-4)` mol
For a solution having pH=4 , `[H_3O^+]=10^(-4)` M
Conc. of `H_3O^+` in 100 mL = `10^(-4)/1000xx100`
`=10^(-5)` mol
Total `H_3O^+` moles = `4xx10^(-4)+10^(-5)`
`=(4+0.1)xx10^(-4)`
`=4.1xx10^(-4)` mol
Total volume = 400 +100 =500 mL
`[H_3O^+]=(4.1xx10^(-4))/500xx1000`
`=8.2xx10^(-4)` M
`therefore pH=-log (8.2xx10^(-4))`
=4-0.9138=3.0862
(c ) Conc. of `OH^-` in 200 mL of 0.1 M NaOH
`=(0.1xx200)/1000`=0.02 mol
Conc. of `OH^-` in 300 mL of 0.2 M KOH =`(0.2xx300)/1000`=0.06 mol
Total moles of `OH^-`=0.02+0.06=0.08 mol
Total volume =200+300 =500 mL
`[OH^-]=0.08/500xx1000`=0.16 M
pOH=-log (0.16)=0.796
`therefore` pH=14-0.796=13.204