Chemistry calculation problems

Question

1. 10,0 cm^3 0,1 mol/l solutions of HCl and CH3COOH are given
2. What amount x cm^3, x mol/l NaOH is necessary to neutralize both solutions
3. 10,0 cm^3 0,001 mol/l HCl was diluted to 1 l. Find the pH of the obtained solution.
4. Prove that after mixing the same volume 0,25 mol/l HCl and 0,25 mol/l NaOH solutions, we get the solution where joint number of ions is half smaller than before the mixing
5. Al2(SO4)3. Find the molar concentration of Al if molar concentration of sulfate is 0,30 mol/l

Answer 1

1). NaOH + HCl ==> NaCl + H₂O

NaOH + CH₃COOH ==> CH₃COONa + H₂O

In both cases the mole ratio of NaOH to acid is 1 : 1, so in each case it will take 10.0 cm³ of 0.1 mol/L NaOH solution

2). (10.0 cm³)(0.001 mol/L) = (1000 cm³)(x mol/L)

x = 0.00001 mol/L = 1x10^-5 M

pH = - log [H+] = - log 1x10^-5

pH = 5

3). Mixing equal volumes of two solutions dilutes each solution by half, so number of ions will be 1/2 the original. Not really sure what the question is getting at. It’s a pretty foolish question IMHO.

4). Al₂(SO₄)₃ ==> 2Al³⁺ + 3SO₄^2-

0.3 mol/l SO₄^2- x 2 mol/l Al³⁺ / 3 mol/l SO₄^2- = 0.2 mol/l Al³⁺