Two 20.0 mL samples, one 0.200 MKOH and the other 0.200 M CH3NH2, were titrated
ID: 969978 • Letter: T
Question
Two 20.0 mL samples, one 0.200 MKOH and the other 0.200 M CH3NH2, were titrated with 0.100 MHI. Answer each of the following questions regarding these two titrations.
What is the volume of added acid at the equivalence point for KOH?
What is the volume of added acid at the equivalence point for CH3NH2?
Predict whether the pH at the equivalence point for each titration will be acidic, basic, or neutral.
Predict whether the at the equivalence point for each titration will be acidic, basic, or neutral.
neutral for both
Predict which titration curve will have the lowest initial pH.
Predict which titration curve will have the lowest initial .
neutral for KOH and acidic for CH3NH2 neutral for KOH and basic for CH3NH2 acidic for KOH and neutral for CH3NH2 basic for KOH and neutral for CH3NH2neutral for both
Predict which titration curve will have the lowest initial pH.
Predict which titration curve will have the lowest initial .
The initial pH will be lower for CH3NH2. The initial pH will be lower for KOH.Explanation / Answer
no of moles of KOH = molarity * volume in L
= 0.2*0.02 = 0.004 moles
at equivalent point no of moles of acid and no of moles of base are equal
no of moles of HI = no of moles of KOH
no of moles of HI = 0.004 moles
Molarity of HI = no of moles of HI/volume in L
0.1 = 0.004/volume in L
volume in L = 0.004/0.1 = 0.04 L =40ml
2, no of moles of CH3NH2 = molarity * volume in L
= 0.2*0.02 = 0.004 moles
at equivalent point no of moles of acid and no of moles of base are equal
no of moles of HI = no of moles of CH3NH2
no of moles of HI = 0.004 moles
molarityof HI = no of moles of HI/volume in L
0.1 = 0.004/voluem in L
volume of HI = 0.004/0.1 = 0.04 L
pH at the equivalence point KOH is 7 ( strong base + Strong acid)
pH at the equivalence point oc CH3NH2 is PH<7 ( weak base + strong acid)
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