Question

2. 200 ml of 0.1 M Na2CO3 + 100 ml of 0,365% HCl

3. 50ml of 0.1M Na3PO4 + 100ml of 0.05 M H2SO4

4. 80 ml of 0.01M HNO2 + 20ml of 0.05 M

HNO3

5. 20 ml of 0.05 M H2C2O4 + 200 ml of 0.01 M KOH

6. 100 ml of 0.05 M H2S + 100 ml of 0.1 M NaOH

How to find pH of solutions

Answer 1

To calculate pH of 80ml 0.01M HNO2 + 20ml of 0.05M HNO3:

Each acid will dissociate to give H+ ions.

pH is defined as the negative logarithm of H+ion concentration present in the solution.

All acids have pH < 7

water or any neutral solution has pH =7

All bases have pH>7

As we got a mixture of two different acids here, we will get only H+ ions concentration in the solution and pH <7

Lets calculate the number of moles of H+ ions from each acids

generally the formula to calculate molarity is

molarity = numberofmoles/volumesofthesolutioninlitres

so number of moles = molarity x volume of the solution in litre

From HNO2, number of moles of H+ ions = (0.01x80)/1000 = 0.8 x 10 ^ -3

From HNO3, number of moles of H+ ions = (0.05x20) / 1000 = 1.0 x 10^ -3

Total number of moles of H+ ions = 1.8 x 10^-3

Total volume of the solution = 80 + 20 = 100 ml

Hence, the net H+ ion concentration = (1.8 x10^-3) x1000 /100 = 1.8 x 10^-2

Therefore , pH = - [H+]

= - [ 1.8x10^-2] = 1.745 ...... answer