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 empirical formula of a compound containing 77.7% Iron and 22.3% of oxygen.
Explain the trend in melting points of the period 3 elements
Explain, in the context of catalysis, the term heterogeneous and describe the first stage in the mechanism of this type of catalysis.
Write down the equation for the Gibbs Free Energy change of a reaction. Hence explain why, for a spontaneous endothermic reaction, there must be an increase in the total entropy.