so the amount of acid in the solution is 33.7 mink Because this is equivalent to

Question

so the amount of acid in the solution is 33.7 mink Because this is equivalent to the 230 grains of unknown acid we started with, we write 2.50 g acid 33.7 mmol acid = 3.37 x 10^-2 mol acid Dividing by 3.37 x 10^-2 gives 74.2 g acid 1.00 mol acid Thus, the formula mass of the unknown acid is 74.2. PRACTICE PROBLEM 12-10: A 2.50-gram sample of malonic acid is dissolved to make 100 mL of solution. The solution is neutralized with 28.5 ml of 1.684 M KOH(aq). Given that the formula mass of malonic acid is 104.1, determine how many acidic protons it has per formula unit. Answer: two The following Example and Practice Problem incorporate many of the concepts that we have learned so far. EXAMPLE 12-11: Calculate the concentrations of all the ions in solution following the reaction of 25.0 ml of 0.200 M CaCl2(aq) with 25.0 ml of 0:300 M AgNO3(aq). Solution: Applying the solubility rules (Table 10.9), we find that the reaction produces an insoluble silver chloride. AgCI(s), precipitate. The balanced chemical equation for this reaction is

Explanation / Answer

1. suppose there are n acidic protons.

then the solution is n*2.50*1000/(104.1*100) normal = 0.240*n N

so, (0.240*n)*100 = (1.684)*28.5 ..... (N1V1 = N2V2)

so, n = 1.99 ~ 2

2. Before Reaction

mols of CaCl2 = 0.2*25 mMols = 5 mMols i.e. Ca2+ = 5 mMol, Cl- = 10 mMol

mols of AgNO3 = 0.3*25 mMols = 7.5 mMols i.e. Ag+ = 7.5 mMol = NO3-

after reaction

AgCl will precipitate. Here, limiting reagent is AgNO3.

So, 7.5 mMol AgCl precipitates.

Remaining ions: Ca2+ = 0.005 mol, Cl- = (10-7.5)*10^-3 mol = 0.0025 mol, NO3- = 0.0075 mol

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