7.49 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) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

(a)

For 2g of TIOH dissolved in water to give 2 L of the solution:

[TIOH(aq)]=22g/L=22×1221M=1211MTIOH(aq)TI(aq)++OH(aq)-[OH(aq)-]=[TIOH(aq)]=1211MKw=[H+][OH-]10-14=[H+]1221221×10-14=[H+]pH=-log[H+]=-log(221×10-14)=-log(2.21×10-12)=11.65

(b)

For 0.3 g of CaOH2 dissolved in water to give 500 mL of solution:

Ca(OH)2=0.3×1000500=0.6M[OHaq-]=2×[Ca(OH)2aq]=2×0.6= 1.2 M[H+]=Kw[OHaq-]=10-141.2M=0.833×10-14pH=-log(0.833×10-14)=-log (8.33×10-13)=(-0.902+13)=12.098

 

(c)

For 0.3 g of NaOH dissolved in water to give 200 mL of solution:

NaOHNa(aq)++OH(aq)-[NaOH]=0.3×1000200=1.5M[OHaq-]=1.5MThen,[H+]=10-1\41.5= 6.66×10-13pH=-log(6.66×10-13)=12.18

(d) For 1mL of 13.6 M HCl diluted with water to give 1 L of the solution:

13.6 × 1 mL = M2 × 1000 mL

(Before dilution) (After dilution)

13.6 × 103 = M2 × 1L M2

1.36 × 10-2 [H+] = 1.36 × 10-2 pH =  log (1.36 × 10-2)

= (– 0.1335 + 2) = 1.866 = 1.87