Calculate the pH of the following solutions:

(a) 5.00 g of HBr in 100mL of aqueous solution

(b) 1.50 g of NaOH in 50mL of aqueous solution

MolarmassofHBr=1+80=81g/mole

Number of moles of HBr = Givenweight = 5 = 0.06173 moles

Molar mass 81

Molarity of HBr = Numberofmoles = 0.06173 = 0.6173 M

Volume of solution in L 0.10

HBrdissociatesinsolutionas:

HBr→H⁺+Br^-

Concentration of HBr = Concentration of H⁺ = 0.6173 M

pH = - log [H⁺]

=-log(0.6173)=0.21

So, pH of HBr is 0.21

Molar mass of NaOH = 23+16+1 = 40 g/mole

Number of moles of NaOH =Given weight = 1.5 = 0.0375moles

Molar mass 40

Molarity of NaOH = Numberofmoles = 0.0375 = 0.75 M

Volume of solution in L 0.050

NaOHdissociatesinsolutionas:

NaOH → Na⁺ + OH^-

ConcentrationofOH^-- =0.75M

pOH = - log [OH⁺]

=-log(0.75)=0.1249

pH = pK_w – pOH = 14.0 – 0.1249 = 13.88

So, pH of NaOH is 13.88