What is the oh- in the final solution prepared by mixing 20 ml of 0.050 m hcl?
Answers
Answered by
3
Explanation:
umber of milliequivalents of HCl = 2 x 0.050 x 1 = 1Number of milliequivalents of Ba(OH)2 = 2 x 30 x 0.10 = 6[OH]- of final soluton ..
Answered by
0
Explanation:
ANSWER
Ba(OH)
2
+2HCl⟶BaCl
2
+2H
2
O
No. of moles of H
+
ions present in solution = concentration of HCl× Volume of HCl
⇒0.05×
1000
20
=0.001 moles
There are two OH
−
ions in one molecule of Ba(OH)
2
.
OH
−
ions released in the Barium Hydroxide solution = concentration of Ba(OH)
2
× Volume of Ba(OH)
3
×2
⇒[OH
−
]=0.1×
1000
30
×2=0.006 moles
∵H
+
ion is limiting reagent.
∴ 0.001 moles of H
+
and OH
−
combine to form water.
∴ 0.005 moles of OH
−
ions are remaining.
Total volume of solution =(20+30)mL=50mL=0.05L
∴ The molarity of OH
−
ion=
0.05L
0.005 moles
=0.100M
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