Urgent Chemistry help! When 0.1009 g of sodium cyanide is dissolved in solution, it takes 19.22 mL...

When 0.1009 g of sodium cyanide is dissolved in solution, it takes 19.22 mL of a potassium permanganate solution to reach the end point of the titration. Using the balanced chemical equation provided, determine the molarity of the potassium permanganate. (Remember: the mole ratio of MnO4- to KMnO4 is 1:1 and the mole ratio of CN- to NaCN is 1:1).

H2O (l) + 2 MnO4- (aq) + 3CN- (aq) ------> 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)

I am completely lost and have no idea what to do. If anyone can provide a step-by-step, explained answer it would be greatly, greatly appreicated. Thanks a ton.

It's the mole ration between CN- and MnO4- that counts since those are the two reacting.

Convert the 0.1009 g of NaCN into moles Since the mole ration between Na+ and CN- is 1:1 then finding the number of moes of NaCN will also give you the number of moles of CN- reacting.

n = m/M

n = 0.1009g / 49.01 g/mol = 0.002058 mol

The mole ratio in the equation between MnO4- and CN - is 2 : 3 Than means that for every 1 mole of CN- you have only 2/3 mol of MnO4- is needed.

Therefore multiply the number of moles of CN- you have by 2/3

0.002058 x 2/3 = 0.00137 mol of MnO4- needed.

Now that you have the moles of MnO4- and the volume of MnO4- used you can calculate the concentration of the MnO4- by using C = n/V

C = 0.00137 / 0.01922 L = 0.0714 mol/L