Question

2) In preparing the KHP solution, a student did find a 100 ml volumetric flask and instead used a 200 mL size to dissolve the same weight of solid to be dissolved in 100 mL solution. In comparison to preparing a 100 mL solution as described in the procedure, explain how the student's action will affect (if any): a) The number of moles of KHP in a 25.00 mL aliquot of the solution. b) The equivalence point of the titration? c) The stoichiometry of the reaction?

Answer 1

answ ! standard solution preparation always use fox size of volumetric which are made on iso standard value. so uncertainity seduce. deviation less. compariso of beth we prepare two solution of kHp in 100ml and 200ml and we find out difference in both of them. solution preporation :- Weight is not given (if given in your priveous. either in standarteon, use providede weight at the weight position, method always same only value change) so weruse we use Imole weight of KAPE - molar mass of K HP < 04.22 gm mols, and dis solve in 200ml we find out concentration of solution. Now molarity = 204.22 x 1000 - 200.22 x 200 also written I molen x 1000 200 MI : 54

room than molarity Now if we take e mole x 1000100 10M them solution. both of twice Difference becomes s We use 25ml aliquet than no. of mole : 5m x orm! 125 mil mole. make in 100 ml than) 25 ml aliquot than We use com culhich 2 • 250 milimole equivalent point both have same mole we also use equation for equality KHP SIDE C1V1 : 62v2 NAOH SIDE E Example = 5m x 25 mL = (2x v2 : 125 Molomole both side. 31 stichiometry KHC84404 + NAOH ---> H2O + NARC8H404 Mole yatio : KHP : NAOH.