You create solutions of HI and NaOH with concentrations of 1.12 and 0.85,respect

Question

You create solutions of HI and NaOH with concentrations of 1.12 and 0.85,respectively. If you titrate 10.0 mL of the HI solution with the NaOH base you have created, at what volume do you expect to see the equivalence point?
mL NaOH = ???
If the actual mL used is 16, what was the actual concentration of the base assuming the acid concentration was correct? Actual [NaOH] =  ??? M
If the actual mL used is 16, what was the actual concentration of the acid assuming the base concentration was correct? Actual [HI] = ???   M
What is the %error (+ or -) in each case?
%Err = (Actual - Given)/Actual x 100%
%error [NaOH] = ???
%error [HI] = ???

Explanation / Answer

HI + NaOH ----> NaI + H2O

M1V1 = M2V2

1.12*10 = 0.85*V2

V2 = 13.176 ml of NaOH at equivalence point

Then

1.12*16 = 13.176*M2

M2 = 1.36 M ---NaOH

Then

M1*16 = 13.176*0.85

M1 = 0.7 M HI

% error [NaOH] = 37.5

% erroe [HI]= -60

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.