Calculate the pH of the following solutions :
(a) 2 g of TIOH 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 diluted with water to give 1 litre of solution.

Explanation

(a) pH of 2 L solution of 2g TlOH :
Molecular mass of TlOH =204+16+1=221 g mol
Molarity of TlOH =`"Weight"/"Molecular mass x Litre"`
`=2/(222xx2)=4.525xx10^(-3)` M
TlOH is strong base, complete ionization occurs so in solution,
`[OH^-]=[TlOH] = 4.525xx10^(-3)` M
pOH=-log `(4.525xx10^(-3))`
`=-[log 4.525+log 10^(-3)]`
=-(0.6556-3.0)=2.3444
pH = 14.0 - pOH = 14.0 - 2.3444
= 11.6556 `approx` 11.66
(b) pH of 500 mL solution of 0.3 g `Ca(OH)_2` :
Molecular mass of `Ca(OH)_2`=40+(16+1)2
= 74 g `mol^(-1)`
`M="Weight in g"/"Molecular mass x Volume"`
`=((0.3g))/((74 "g mol"^(-1))(0.5L))=8.108xx10^(-3)` M
`Ca(OH)_2` is strong , 100% dissociation of base,
`Ca(OH)_2 to Ca_((aq))^(2+) + 2OH_((aq))^(-)`
`therefore` In solution `[OH^-]=2xx[Ca(OH)_2]`
`=(2xx8.108xx10^(-3))M`
`=16.216xx10^(-3)` M
pOH=log `[OH^-]`=-log `(16.216xx10^(-3))`
=-(1.2099-3.0)=-(-1.7901)
=+1.7901
pH = 14.0 -pOH
=14.0-1.7901=12.2099 `approx` 12.21
(c) pH of 200 mL solution of 0.3 g NaOH :
Molarity = `"Weight x 1000"/"Molecular mass x Volume mL"`
`=(0.3xx1000)/(40xx200)=(0.3 g xx 1000)/((40 g "mol"^(-1))xx(200 mL))`
`=3.75xx10^(-2)` M
Strong base NaOH complete ionization in solution,
`[OH^-]=[NaOH ]=3.75xx10^(-2)` M
`pOH=-log [OH^-]=-log (3.75xx10^(-2))`
=-(0.5740-2.0)=-(-1.426)=1.426
pH = 14.0-pOH =14.0 - 1.426 = 12.574 `approx` 12.57
(d) pH of 1 L solution made in 1 mL of 13.6 M HCl :
`{:("Initial stage","After dilution"),(M_1V_1=,M_2V_2):}`
`therefore 13.6xx1 mL = M_2xx1000 mL`
`therefore M_2=1.36xx10^(-2) M HCl =[H^+]`
`pH = -log [H^+]- - log (1.36xx10^(-2))`
=2-0.1335= 1.8665 `approx` 1.87