1.025g of an impure sample of a weak acid HX(mol mass = 82) is dissolved in 60ml of water and titrated
with 0.25M NaOH. When half of the acid was neutralized the pH was found to be 5.0 and at the equivalence
point the pH is 9.0. Find the % purity of this sample of HX
Pls help me doing this question
Frankly I think this problem is lacking some information. As an estimate this is what I would do.
pH at the half way point is 5.0 which means the pKa of HX is 5.0 and that makes Ka = 1E-5.
At the eq point you have the hydrolysis of X^- (the anion) as
.....X^- + HOH ==> HX + OH^-
I....y.............0.....0
C...-z.............z.....z
E...y-z............z.....z
The problem tells you that the pH at the eq point is pH = 9. That makes pOH = 5 because pH + pOH = pKw = 14. So (OH^-) = 1E-5.
Kb for X = (Kw/Ka for HX) = (z)(z)/(y-z)
You know Kw, Ka, and z which allows you to calculate y which = (HX) and my estimate of that is 0.1M but you should go through the math and confirm that. That is 0.1 mol/L. You have it in 60 mL. Convert that to mols in the 60 mL, multiply by 82 to find grams in the 1.025 g sample.
Then % HX = (grams HX/1.025)*100 = ? Post your work if you get stuck.
NOTE: The reason I think this is an estimate(i.e., something is missing in the problem) is because when you calculate (HX) at the equivalence point, that is (HX) AFTER the 0.25 M NaOH has been added so the volume at that point is the 60 mL you started with + the amount of 0.25M NaOH added and I don't see any way to estimate how much of the 0.25M NaOH is used. Therefore, I don't think an unqualified answer for % purity can be given. Perhaps another tutor will see something I don't see.
Thank you DrBob222...
Got the answer as 80% purity.
To find the % purity of the sample of weak acid HX, we need to determine the moles of acid present, as well as the moles of impurities.
Step 1: Calculate the moles of NaOH used
To find the moles of NaOH used for neutralization, we need to use the given concentration and volume of NaOH used during the titration. The formula to calculate moles is:
Moles = Concentration (M) x Volume (L)
In this case, the concentration of NaOH is 0.25 M, and the volume used is half of the total volume at equivalence point, so 60 ml / 2 = 30 ml = 0.03 L.
Moles of NaOH used = 0.25 M x 0.03 L
Step 2: Calculate the moles of HX at equivalence point
The moles of NaOH used in the reaction are equal to the moles of HX at the equivalence point, as they react in a 1:1 ratio.
Moles of HX at equivalence point = Moles of NaOH used
Step 3: Calculate the moles of HX in the impure sample
The moles of HX at the equivalence point represent the total moles of acid present, including impurities. However, at the half-equivalence point, only half of the acid is neutralized.
Moles of HX in the impure sample = Moles of HX at equivalence point / 2
Step 4: Calculate the moles of impurities
The moles of impurities can be obtained by subtracting the moles of HX in the impure sample from the moles of HX at the equivalence point.
Moles of impurities = Moles of HX at equivalence point - Moles of HX in the impure sample
Step 5: Calculate the moles of pure HX
The moles of pure HX can be obtained by subtracting the moles of impurities from the moles of HX in the impure sample.
Moles of pure HX = Moles of HX in the impure sample - Moles of impurities
Step 6: Calculate the mass of pure HX
To calculate the mass of pure HX, we use the formula:
Mass (g) = Moles x Molar mass
In this case, the molar mass of HX is given as 82 g/mol.
Mass of pure HX = Moles of pure HX x Molar mass of HX
Step 7: Calculate the % purity of the sample
The % purity of the sample can be calculated by using the formula:
% Purity = (Mass of pure HX / Mass of the impure sample) x 100
Now you can substitute the values calculated in the previous steps to find the % purity of the sample.