Titration Part. Use isocyanic acid from problem 5 in this titration. You may need its Ka. A 25.00 mL solution of 0.0300 M isocyanic acid is titrated with 0.0400 M NAOH solution. What volume of added base do you expect the equivalence point? HOCN (aq) + NaOH (aq) → H;0 (1) + Na* OCN- (aq) What is the volume of the solution at the equivalence point (assuming that the volumes of the two solutions are additive)

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6ab) please see attached

Ka=2.39x10^-4

Titration Part. Use isocyanic acid from problem 5 in this titration. You may need its Ka.
A 25.00 mL solution of 0.0300 M isocyanic acid is titrated with 0.0400 M NAOH solution. What volume of
added base do you expect the equivalence point?
HOCN (aq) + NaOH (aq) → H;0 (1) + Na* OCN- (aq)
What is the volume of the solution at the equivalence point (assuming that the volumes of the two solutions
are additive)
Transcribed Image Text:Titration Part. Use isocyanic acid from problem 5 in this titration. You may need its Ka. A 25.00 mL solution of 0.0300 M isocyanic acid is titrated with 0.0400 M NAOH solution. What volume of added base do you expect the equivalence point? HOCN (aq) + NaOH (aq) → H;0 (1) + Na* OCN- (aq) What is the volume of the solution at the equivalence point (assuming that the volumes of the two solutions are additive)
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