Calculate the pH of a 0.0131 mol dm^-3 solution of calcium hydroxide at 10 degrees centigrade.
Multiply by 2 because calcium hydroxide = Ca(OH)2 so 2 x [OH-] per molecule.[OH-] = 0.0131 x 2 = 0.0262
Insert [OH-] value into the equilibrium equation along with the value of Kw at 10 degrees centigrade 2.93 x 10^-15 (from data tables). This gives the value for [H=].[H+] = (Kw/[OH-] ) = 2.93 x 10^-15 / 0.0262 = 1.118 x 10^-13
Finally, insert [H+] value into pH equation.pH = -log (1.118 x 10^-13) = 12.95
EW
Related Chemistry A Level answers
All answers ▸Calculate the pH change when water is added to 25.0 ml of 0.250M NaOH to form a 1.00l solution.
Describe a test that could be used to distinguish between potassium sulfate (K2SO4) and potassium nitrate (KNO3)
Calculate the empirical and molecular formula of the molecule giving rise to the molecular ion peak at 148 m/z. The percentage composition by weight is 64.80 % carbon, 13.62 % hydrogen, and 21.58 % oxygen
Why is there a difference in mechanism between tertiary halogenoalkanes and primary halogenoalkanes in nucleophilic substitution?