Calculate the `pH` of the following solutions:
a. `2 g` of `TlOH` dissolved in water to give `2` litre of solution.
b. `0.3 g` of `Ca(OH)_(2)` dissolved in water to give `500 mL` of solution.
c. `0.3 g` of `NaOH` dissolved in water to give `200 mL` of solution.
d. `1 mL` of `13.6 M HCl` is duluted with water to give `1` litre of solution.

Explanation

Molar concentration of the solution
`=("Amount of TlOH//litre of solution")/("Molar mass of TlOH")`
`=1/221 M=0.0045`
`[H_(3)O^(o+)]=K_(w)/[overset(Θ)(O)H]=10^(-4)/0.0045=10^(-10)/45`
`pH=-log[H_(3)O^(o+)]=- log[10^(-10)/45]`
`=-[-10 log 10 -log 45]=-[-10-1.6532]`
`=10+1.6532=11.66`
Alternatively:
`[overset(Θ)(O)H]=45xx10^(-4)M`
`pOH=- log(45xx10^(-4))=-log 3(2)-log 5+4`
`=-2xx0.48-0.7+4=2.34`
`pH=14-2.34=11.66`
b. Molar concentration of the solution
`=("Amount of" Ca(OH)_(2)//litre of solution)/("Molar mass of" Ca(OH)_(2))`
`=0.6/74=0.008 M`
`Ca(OH)_(2)rarr Ca^(2+)(aq)overset((2xx0.008M))(+2overset(Θ)(O)H)`
`[overset(Θ)(O)H]=2xx8xx10^(-13)=16xx10^(-3)`
`[H_(3)O^(o+)]=K_(w)/([overset(Θ)(O)H])=(10^(-14)M^(2))/(16xx10^(-3))=1/16xx10^(-11) M`
`pH=-log [H_(3)O^(o+)]`
`=-log[1/16xx10^(-11)]=[log 1-4 log 2-11]`
`=-[-4xx0.3010-11]=-[-12.2040]=12.20`
Alternatively:
`[overset(Θ)(O)H]=16xx10^(-3)M`
`pOH=-log 2^(4)+3=-1.2+3=1.8`
`pH=14-1.8=12.20`
c. Molar concentration of the solution
`=("Amount of" NaOH//litre of solution)/("Molar mass of NaOH)`
`=1.5/40 M=0.037 M`
`[overset(Θ)(O)H]=0.037 M`
`[H_(3)O^(o+)]=K_(w)/([overset(Θ)(O)H])=((10^(-14)M^(2)))/((0.037))`
`=10^(-14)/(37xx10^(-3))=1/37xx10^(-11)`
`pH=-log[H_(3)O^(o+)]=- log[1/37xx10^(-11)]`
`=-[(log)1/37-11]=-[log 1- log 37-11]`
`=-[-1.5682-11]=12.5682=12.57`
Alternatively:
`[overset(Θ)(O)H]=0.037 M`
`pOH=-log(37xx10^(-3))`
`=-1.5682+3=1.4318~~1.43`
`pH=14 -1.43 =12.57`
d. `M_(1)V_(1)=M_(2)V_(2)`
`13.6xx1=M_(2)xx1000`
`13.6/1000=M_(2)`
or `M_(2)=13.6xx10^(-3)M`
`[H_(3)O^(o+)]=13.6xx10^(-3)M`
`pH=-log [H_(3)O^(o+)]=-log(13.6xx10^(-3))`
`=-[log 13.6-3]=-1[1.1335-3]=-[-1.8665]=+1.8665=1.87`