H2SO4 ---> 2H+ + SO42-
KOH ---> OH- + K+
Initial moles of Sulfuric acid:
Volume x Concentration = Moles
(30/1000) x 0.150 = 4.5x10-3 mol
Moles of H+ made by sulfuric acid:
It is a 1:2 ratio of sulfuric acid to H+ therefore multiply the moles of initial sulfuric acid by 2 to get moles of H+.
(4.5x10-3) x 2 = 9x10-3 moles of initial H+
Initial moles of potassium hydroxide:
Volume x Concentration = Moles
(30/1000) x 0.200 = 6x10-3 mol
Moles of OH- made by potassium hydroxide:
1:1 ratio of potassium hydroxide to OH- therefore 6x10-3 moles of initial OH-
The initial moles of H+ and initial moles of OH- can react to form water leaving behind an excess of H+ ions because there are more of them:
H+ + OH- ---> H2O
(9x10-3) - (6x10-3) = 3x10-3 moles of H+ remaining
pH = -log10([H+]) therefore the new concentration of H+ need to be worked out:
Concentration = Moles / Volume
(3x10-3) / (60/1000) = 0.05 moldm-3 of H+ in solution
then you can plug this into the equation for pH: pH = -log10(0.05) = 1.30.
pH of solution is 1.30