chemistry lab - acid base titration

Trial 1
Molarity of NaOH: 0 .29! M
Volume of NaOH used to reach equivalence poi t
(ml) n q~97m2 41%“ 4.21mi,
L
._ (Marks
51-12%”); 2 )
Average volume of NaOH used to reach equivalence
point (L)
1. What is the number of moles of NaOH needed to reach the equivalence point? (2 Marks) (Show
calculations) M 5 Vi M '7 V 2 M01“
" ‘ '3
V 0~29I Ma+ NmOH x 422““ = 9'Wl0md
2. What is the number of moles of HCl? (2 Marks)
(Show calculations)
mamsmmés mafia
mammals; 35.4%ng :0-94gm0l «we;
Molar! macs 0+ no: 3364693");
20-0 ml of 410120.003
H
7’ A 3; What is the concentration of HCl? (2 Marks)
(Show calculations) 90 ‘ 00 mp/ ”'19 L-
091‘, = a-FAKmoleA 002',
l 0-02L