Help with this ques thanks
0.132g of pure carboxylic acid (R-COOH) was dissolved in 25ml of water and titrated with 0.120mol/L NaOH solution
A volume of 14.80ml was required to reach the endpoint of the titration.
The carboxylic acid could be:
A)HCOOH
B)CH3COOH
C)C2H5COOH
D)C3H7COOH
ans:C
question 2: Oxalic acid C2O4H2
A 25.0ml solution of oxalic acid reacts completely with 15.0ml of 2.50mol/L of NaOH
what is the concentration of the oxalic acid solution?
ans: 0.750M
For Q2, oxalic acid is indeed, diprotic.
2 NaOH
(aq) + C
2O
4H
2(aq) → Na
2C
2O
4(aq) + 2 H
2O
(l)Hence the mole ratio is given
n(NaOH) = 2n(C
2O
4H
2)
Thus, using n = CV
[NaOH] * V(NaOH) = 2* [C
2O
4H
2] * V(C
2O
4H
2)
2.50 mol L
-1 * 15.0 mL = 2 * [C
2O
4H
2] * 25.0 mL
[C
2O
4H
2] = 0.75 mol L
-1 as the answers requested
______________________________________
Q1 is one of those typical rigorous questions that the HSC puts in, where you essentially have to do up to 4 calculations to find the answer.
We wish to determine if methanoic acid (formic acid), ethanoic acid (acetic acid), propanoic acid or butanoic acid was used in the titration. We firstly note that ALL of the carboxylic acids are monoprotic.
Acid + NaOH
(aq) → Sodium salt + H
2O
(l)In every case, the mole ratio is 1 to 1.
Thus, we have
n(Acid) = n(NaOH)
We will apply n=CV to the sodium hydroxide, but n=m/M to the acid:
m/M = CV
0.132g / M = 0.0148 L * 0.120 mol L
-1M = 74.32432... g mol
-1Now that we know the molar mass, we can simply use the periodic table to find out which acid is the correct acid.
It turns out that propanoic acid is correct:
M = 2*12.01 + 5*1.008 + 12.01 + 16.00 + 16.00 + 1.008 = 74.078
The tiny amount of inaccuracy should've been anticipated since the start.