Question

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

Answer 1

a) Tloh is a base,

$[OH] =\frac{n}{v}=[OH^-]= \frac{\frac{2}{204+16+1}}{2}\\ =4.5\times 10^{-3}moles$

$POH=-log(4.5\times 10^{-3})=2.344$

PH = 14-2.344 = 11.65

b) Ca(OH)2 is a base,

POH = $-log(OH^-)$

$\frac{n}{v}=[OH^-]= \frac{\frac{0.3}{40+32+2}}{0.5}\\ =0.0081moles/L$

POH = -log(0.0081) = 2.09

PH = 14-POH =11.91

c) NAOH is a base,

$\frac{n}{v}=[OH^-]= \frac{\frac{0.3}{23+16+1}}{0.2}\\ =0.03751moles/L$

POH = -log(0.0375) = 1.426

PH = 14 - POH = 12.68

d) HCL is a strong acid,

$[H^+]=\frac{n}{v}$

$n =[H^+]\times v =13.6\times 10^{-3}=1.36\times10^{-2}mol/L$

$New [H^+]=\frac{1.36\times 10^{-2}}{1L}$

pH = -log([tex]{1.36\times 10^{-2}}[\tex]) = 1.88