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I think I have a quicker, easier way to solve this problem than the one SaillinX suggested.
Well,
we know we used 20mL OH- Plus! 20mL H+ for a complete neutralization, so...40mL total volume. Judging by the pH, it is obvious, more acid was added than the base, so a little more than 40mL were added so. Hence,
M1V1=M2V2
M1=1M HCL
V1= X HCL
M2= pH = 2 =1*10^-2M Molarity of System.
V2= (40mL + X) Volume of System.
(1M)(V1)=(10^-2)(40mL + V1)
Solve of V1 and you get V1= .3999999 = .4, so Volume of HCl added initially is 20+.4 = 20.4.
I hope this helps
Well,
we know we used 20mL OH- Plus! 20mL H+ for a complete neutralization, so...40mL total volume. Judging by the pH, it is obvious, more acid was added than the base, so a little more than 40mL were added so. Hence,
M1V1=M2V2
M1=1M HCL
V1= X HCL
M2= pH = 2 =1*10^-2M Molarity of System.
V2= (40mL + X) Volume of System.
(1M)(V1)=(10^-2)(40mL + V1)
Solve of V1 and you get V1= .3999999 = .4, so Volume of HCl added initially is 20+.4 = 20.4.
I hope this helps