The volume (in mL) of 0.242 M NaOH solution needed for the titration reaction is 39.44 mL
<h3>Balanced equation </h3>
CH₃CH₂COOH + NaOH —> CH₃CH₂COONa + H₂O
From the balanced equation above,
- The mole ratio of the acid, CH₃CH₂COOH (nA) = 1
- The mole ratio of the base, NaOH (nB) = 1
<h3>How to determine the volume of NaOH</h3>
- Volume of acid, CH₃CH₂COOH (Va) = 46.79 mL
- Molarity of acid, CH₃CH₂COOH (Ma) = 0.204 M
- Molarity of base, NaOH (Mb) = 0.242 M
- Volume of base, KOH (Vb) =?
MaVa / MbVb = nA / nB
(0.204 × 46.79) / (0.242 × Vb) = 1
Cross multiply
0.242 × Vb = 0.204 × 46.79
Divide both side by 0.242
Vb = (0.204 × 46.79) / 0.242
Vb = 39.44 mL
Thus, the volume of NaOH needed for the reaction is 39.44 mL
Learn more about titration:
brainly.com/question/14356286
Answer:
yh
Explanation:
welcome.This place is for learning
Answer : The mass of 
is, 295.323 grams
Solution :
First we have to calculate the moles of 
.
Now we have to calculate the moles of NaBr.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 moles of react with 2 moles of
So, 2.87 moles of react with 2.87 moles of
Now we have to calculate the mass of NaBr.
Therefore, the mass mass of is, 295.323 grams
Answer:
Here's what I've found
Explanation:
Refer to the attachment !
