Answered

The initial volume of HCl was 1.25 ml and LiOH was 2.65 ml. The final volume of HCL was 13.60 ml and LiOH was 11.20 ml. If the LiOH was .140 M what was the molarity of HCl ? If the same volumes were used from question above, but the HCl was .140 M , what would the molarity of LiOH be ?

Answer :

13.60-1.25 = 12.35 ml HCl is used
11.20-2.65 = 8.55 ml LiOH is used
HCl+NaOH ---> NaCl +H2O

M - molarity

V(HCl)*M(HCl) =V(LiOH)*M(LiOH)
12.35*M(HCl)=8.55*0.140
M(HCl)=8.55*0.140/12.35=0.097M

V(HCl)*M(HCl) =V(LiOH)*M(LiOH)
12.35*0.140 =8.55*M(LiOH)
M(LiOH)= 12.35*0.140/8.55=0.202 M

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