Find pH of acids

Question

]Calculate the pH of each of the following strong acid solutions: a) 0.0167 M HNO3, b) 0.225 g of HClO3 in 2.00 L of solution, c) 15.00 mL of 1.00 M HCl diluted to 0.500 L, d) a mixture formed by adding 50.0 mL of 0.020 M HCl to 125 mL of 0.010 M HI.

Answer 1

The pH of a strong acid will be the negative log of the concentration

a) -log 0.0167 = 1.78 = pH

b) 0.225 g HClO3 x 1 mol HClO3 / 84.5 g = 0.00266 mols / 2.00 L = 0.00133 M

-log 0.00133 = 2.88 = pH

c) (15 ml)(1.00 M) = (500 ml)(x M) and x = 0.03 M HCl

-log 0.03 = 1.52 = pH

d) 50 ml x 1 L / 1000 ml x 0.020 mol / L = 0.001 moles HCl

125 ml x 1 L / 1000 ml x 0.010 mol / L = 0.00125 moles HI

final volume = 50 ml + 125 ml = 175 ml = 0.175 L

total moles = 0.001 + 0.00125 = 0.00225 moles

concentration = 0.00225 mol / 0.175 L = 0.0129 M

-log 0.0129 = 1.89 = pH