ACID/BASE TITRATION PROBLEM... When hydrogen cyanide (HCN) gas is dissolved in water, it forms a weak acid. the Ka= 6.2 x 10^-10. If a 0.05L sample of 0.1 mol/L HCN (aq) is titrated with 0.025L of 0.1 mol/L KOH (aq), what is the pH of the solution after the base has been added?
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okay so you have KOH yield (I can't make the arrow so it will be complicated) K+ and OH-. It FULLY dissociates, therefore, you will have X amount of mol of OH ( I'll do this in mol but usually I use milimol just for convertion reason) Then you have this reaction HCN + OH- yields H20 + CN - ( conjugate base) (insert amount of moles of HCN ) then amount of moles of OH. The OH ion will reach 0. Therefore Initial amount of HCN - amount of OH will give you the NEW amount of HCN when OH is added. The amount of CN will reach the amount of OH- that was discovered. The point of this problem is that you will have an initial concentration ( new for HCN and for CN) for when you will do the titration reaction. The rest is normal triation. HOWEVER don't forget to find the new concentration using the TOTAL volume. Hope it wasn't a total mess :/
how do you get the amount in mols of OH- in the HCN + OH reaction?
you do KOH yields OH + K. Since k is the conjugate of a strong acid, it doesn't affect the reaction at all. KOH dissolves COMPLETELY ( 100% in water) due to its strong basisity. Therefore, the amount of KOH is the same as the amount of OH (1:1 mole ratio)
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sorry i meant the conjugate acid of a strong base *
do you substitute the concentrations of HCN and CN into the Ka equation to find the concentration of OH-?
nononono first of all NEVER , EVER use Ka with OH, if you have OH you'll use Kb. Now, because you're adding a STRONG base to a WEAK acid, the strong BASE will affect the pH by decreasing the amount of weak acid ( and therefore increasing its CONJUGATE BASE amount because NOTHING IS LOSS NOR CREATED) Therefore, KOH COMPLETELY DISSOLVES in water (amount of mole of KOH is the same as amount of OH) NOW, because there is more HCN than OH, the reaction is still ACIDIC. Therefore HCN + OH YIELDS H2O + CN. In this "table" (which isn't an I.C.E table) , you will bring the OH to 0 and decreasing the amount of HCN (therefore INITIAL MOLE which should be found by its initial volume * concentration) BY THE AMOUNT OF OH THAT WAS INSERTED IN THE SOLUTION. Also BECAUSE HCN decreases, the INITIAL amount of CN MUST increase ( by the same amount because of a 1:1 ratio all across). Afterwards SET UP AN ICE EQUATION with the INITIAL concentration of both HCN and CN|dw:1355796500909:dw| THE NUMBER OF THIS DRAWING OR ONLY HYPOTHETICAL