A contains 2g/200cm³ of H2X solution. B contains 2.85g/250cm³ of KOH solution. If 18.70cm³ of A neutralized 25cm³ of B. Calculate the;
i) Concentration of A in mol/dm³
ii) Molar Mass of acid H2X.
iii) Percentage by mass of X in H2X.
NOTE: I think you meant 2 g H2X/200 cc solution for A and 2.85 g KOH/250 cc solution for B. I will assume that's what you meant.
(A) = (H2X) = 2 g H2X/200 cc x (1000 cc/200) = 10 g H2X/L of solution.
(B) = KOH = 2.85 g KOH/250 cc x (1000 cc/250) = 11.4 g KOH/L of solution.
Find molarity of KOH. mols KOH = g/molar mass = 11.4/56.1 = 0.203 mols/L = 0.203 M for KOH.
The reaction is H2X + 2KOH ==> K2X + 2H2O
cc KOH = 25.0 cc. cc H2X = 18.70 cc.
millimols KOH = mL x M = 25.0 x 0.203 = 5.08
millimols H2X = 1/2 that or 5.08/2 = 2.54 (because it's 1 mol H2X for 2 mols KOH. millimols H2X = mL x M or M = millimols/mL = 2.54/18.70 = 0.135 M for (H2X) = 0.135 mol/dm3
mols H2X = grams/molar mass or molar mass = grams/mols = 10/0.135 = 74.1
(iii) molar mass H2X = 74.1. H = 2*1 = 2 and 74.1 - 2 = 72.1 for X
%X in H2X = (72.1/74.1)*100 = ?
Please check the math thoroughly. I made a couple of math error but I think I caught and corrected all of them. For whatever it's worth I don't know of an anion that has an atomic mass of 72.1 so I assume this is a made up problem.
To solve this problem, we need to use the concept of stoichiometry and the equation for the neutralization reaction between H2X and KOH. The balanced equation for this reaction is:
H2X + KOH → H2O + KX
Now, let's work through each part of the question step by step:
i) Concentration of A in mol/dm³:
Given that 18.70 cm³ of A neutralized 25 cm³ of B, we can determine the ratio of the volumes of A to B.
Volume of A / Volume of B = 18.70 cm³ / 25 cm³ = 0.748
Since the ratio of the volumes is the same as the ratio of the moles, we can conclude that 0.748 mol of A neutralized 1 mol of B.
Using this information, we can calculate the concentration of A in mol/dm³ by dividing the number of moles by the volume in dm³.
Concentration of A = (0.748 mol) / (0.0187 dm³) ≈ 40.00 mol/dm³
ii) Molar Mass of acid H2X:
The molar mass of H2X can be determined by comparing the number of moles of H2X and KOH in the neutralization reaction.
From the balanced equation, we can see that 1 mole of H2X reacts with 1 mole of KOH.
From part i), we know that 0.748 mol of H2X reacts with 1 mol of KOH.
Therefore, the molar mass of H2X is equal to the molar mass of KOH, which is 2.85 g/0.0250 dm³.
Molar mass of H2X = 2.85 g / 0.0250 dm³ ≈ 114 g/mol
iii) Percentage by mass of X in H2X:
To determine the percentage by mass of X in H2X, we need to calculate the molar mass of X and then calculate the percentage.
From the balanced equation, we know that 1 mole of H2X contains 1 mole of X.
The molar mass of X is calculated by subtracting the molar mass of H2 (2 * atomic mass of hydrogen) from the molar mass of H2X.
Molar mass of H2 = 2 * 1.01 g/mol = 2.02 g/mol
Molar mass of X = molar mass of H2X - molar mass of H2 = 114 g/mol - 2.02 g/mol
Now we can calculate the percentage by mass of X:
Percentage by mass of X = (Molar mass of X / Molar mass of H2X) * 100
Percentage by mass of X = (111.98 g/mol / 114 g/mol) * 100 ≈ 98.24%
Therefore, the percentage by mass of X in H2X is approximately 98.24%.