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 ▸When an unsymmetrical alkene undergoes electrophilic addition you often get a major and minor product. What would the major product be when propene reacts with hydrochloric acid? Why is this?
How do you convert between different SI units, for example, nmol to mmol?
give a possible reaction mechanism for the conversion of a haloalkane to alcohol
20cm3 of 0.5moldm-3 of HCL is diluted by adding 15cm3 of water. This diluted solution is titrated against a 0.3moldm-3 solution of NaOH. What is the volume of the NaOH in cm3 required to reach the endpoint of the titration?