Question from surfing acidic environment (1 Viewer)

[let.me.die]

i was wondering if someone could show me how to do this question from the surfing book for acidic environment.
12. Calculations of pH (pg.19)
5. Calculate the pH of the solutions produced by:
(c) mixing 50mL of 0.1 mol L HCl with 20 mL of 0.05 mol L NaOH.
Ans:1.24

 

[Dreamerish*~]

HCl + NaOH → NaCl + H₂O

n_HCl = 0.05 x 0.1 = 0.005
n_NaOH = 0.02 x 0.05 = 0.001

.: HCl is in excess. For each mole of HCl, one mole of NaOH reacts with it. We have 0.005 moles of HCl but only 0.001 moles of NaOH. Hence 0.001 moles of HCl will react with the NaOH, and the other 0.004 moles will be left in solution.

Because NaOH and HCl are both strong, the salt solution produced will be neutral (strong acid + strong base = neutral salt solution). So the only substance that will affect the pH of the salt solution is the HCl left over after neutralisation. This HCl is still in solution, and therefore can ionise (completely) to produce hydronium ions.

After neutralisation, we still have 0.004 moles of unused HCl. This is in 70 mLs of salt solution. So the concentration of H⁺ is 0.004/0.07 = 0.057 M.

pH = -log₁₀[H⁺]

= -log₁₀0.057

= 1.24