50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is
a.
4 g
b.
20 g
c.
8 g
d.
10 g
a.
4 g
b.
20 g
c.
8 g
d.
10 g
Question ID - 52376 | SaraNextGen Top Answer 50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is
a.
4 g
b.
20 g
c.
8 g
d.
10 g
50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is
|
a. |
4 g |
b. |
20 g |
|
c. |
8 g |
d. |
10 g |
1 Answer
127 votes
Answer Key / Explanation : (a) -
H2C2O4 + 2NaOH 
Na2C2O4 + 2H2O
meq of H2C2O4 = meq NaOH
50 × 0.5 × 2 = 25 × MNaOH × 1
MNaOH = 2 M
Now 1000 ml solution = 2 × 40 gram NaOH
50 ml solution = 4 gram NaOH
127 votes
Answer Link
127