Question

Be sure to answer all parts. Find the pH of the equivalence point and the volume (mL) of 0.0702 M KOH needed to reach the equivalence point in the titration of 23.4 mL of 0.0390 M HNO,. Volume: mL KOH pH =

Answer 1

leto vme of 0. 0102 M KOH needed to reach the eguirchent point. 1. Su Visi= Nath VX 0.070 2M = 23.4 mil x 0.0390 - N 0.039 Z 23.4 mL x 0:0 102 13 so, volume of kom needed = 13 ml. so, scut hacholysis As at the the auld is wean ened point. ONOZ + tho +IN02 + ou - 5 o ca a [*2 cow) CA Eng - = ella] As KKRI so lani so Tko = xtc) as a = √ 4 23.4 LX0.039M [NO2- at the end point = 36.4AL : 0025 M you-] = c.x= c. 52 kb Kahno, 4-6710-4 ko = 10-14 - 21174%20- 10-14ka = 6x04 = 2 Sur [ow'] = 10.025 x 2.174 X10-4 = 7.372x10-7 pOH = -log(7-372x10-) = 6.132 PH: 14 - DOH: 14 - 6.132 = 7.868