In: Chemistry
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Solution :-
Given data
15 ml 2.6 M HCl
Part a) 15 ml 2.6 M HCl added to 600 ml pure water
Lets first calculate final concentration of the HCl after mixing the 15 ml 2.6 M HCl with 600 ml pure water
After mixing final volume is = 15 ml + 600 ml = 615 ml
Formula to calculate final concentration is as follows
M1V1=M2V2
Where M1 = initial concentration
V1= initial volume
M2= Final concentration
V2= final volume
Lets put the values in the formula and solve for M2
M1V1=M2V2
2.6 M * 15 ml = M2 * 615 ml
M2 = (2.6 M * 15 ml )/615 ml
M2 = 0.0634 M
HCl is the strong acid so it dissociate completely therefore concentration of the H+ is same as concentration of HCl that is 0.0634 M
Lets calculate pH using this H+ concentration
Formula to calculate pH is as follows
pH= - log[H+]
pH= - log [ 0.0634]
pH= 1.20
part b) 15 ml 2.6 M HCl added to 600 ml 0.10 M CH3COO-
reaction equation is as follows
HCl + CH3COO-