What mass of sodium cyanide must be added to 250ml of water in order to obtain a

Question

What mass of sodium cyanide must be added to 250ml of water in order to obtain a solution having a pH of 10.50? Ka(HCN) = 4.9 x 10 ^-10 Please give a step by step solution. I am really confused with this.

Explanation / Answer

CN- + H2O--> HCN + OH-........... Use the pH to calculate [H+] at equilibrium.......[H+]=10^(-10.50)=3.162*(10^-11) M Use [H+] to find [OH-]...........[OH-]=(10^-14)/[H+]=3.162*10^-4 M Use ka given to find kb for cyanide...........kb=kw/ka=2.04*10^-5...let sodium cyanide be x moles.so cyanide ion will be x moles....at equilibrium [CN-]=x-y;[HCN]=y,..[OH-]=y;...then y=3.162*10^-4;.....[HCN]*[OH-]/[CN-]=kb;on solving we get x=5.2173*10^-3 ;this x is for 1 litre of water for 250 ml number of moles=5.2173*10^-3/4 So mass of NaCN=5.2173*10^-3*(49)/4=0.0639g

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