Please Help! When 0.1009 g of sodium cyanide is dissolved in solution, it takes
ID: 499830 • Letter: P
Question
Please Help!
When 0.1009 g of sodium cyanide is dissolved in solution, it takes 19.22 mL of a potassium permanganate solution to reach the endpoint of the titration. Using the balanced chemical equation provided, determine the molarity of the potassium permanganate. You will be doing a similar reaction in the laboratory this week. (Remember: the mole ratio of MnO_4^- to KMnO_4 is 1:1 and the mole ratio of CN^- to NaCN is 1:1) H2O (l) + 2 MnO_4^- (aq) + 3 CN^- (aq) rightarrow 2 MnO_2 (s) + 3 CNO^- (aq) + 2 OH^- (aq)Explanation / Answer
The endpoint of a titration is the point at which the reaction between the titrant and the analyte becomes complete.
Generally the endpoint of a titration is determined using indicators.
In some cases, either the reactant or the product can serve as the indicator.
A best example is the redox titration using potassium permanganate.
From the above balanced chemical equation, 2 moles of KMnO4 reacts with 3moles of NaCN.
We know the formula :
M1V1/n1 = M2 V2 /n2 ------------------------------------------------(1)
here M1 = molarity of KMnO4 =?
V1= volume of KMnO4 = 19.22ml
n1 = no.of moles of KMnO4=2
M2 = Molarity of NaCN =0.1009g/49g/mol x (1/1L) = 0.00206M
since Molarity = (weight of solute/its molecular weight )x1/volume of solution in litres
volume of NaCN taken for he titration is - 20.0ml
no.of moles of NaCN = 3
hence by substituting all the above values in (1), we get
M1 = M2V2 n1 /n2 V1
= 0.00206M x 20.0ml x 2 / (3x 19.22ml)
= 0.00143M