calculate the pH of a solution obtained by mixing 250 ml of 0.5 M HCL and 750 ml of 0.1 M NAOH
NaOH volume = 750mL or 0.75L
Molarity = 0.1M
thus moles of NaOH = 0.10.75 = 0.075mol [using formula mole = C (concentration in M) V in L]
HCl volume = 250 mL or 0.25 L
molarity = 0.5 M
Moles of HCl = 0.50.25 = 0.125 mol
1 mole of NaOH exactly neutralizes 1 mole of HCl according to the reaction
NaOH + HCl NaCl + H2O
Thus 0.075 moles of NaOH will neutralise 0.075 moles of HCl. Therefore HCl is in excess.
Amount of HCl remaining = 0.125 - 0.075 = 0.05 moles.
Thus concentration of H+ ions =
= 0.05 / 1 ( since volume = 0.25 + 0.75 = 1 L)
= 0.05 M
Thus pH= - log[H+]
where [H+] is concentration of H+ ions in molarity
pH = - log (0.05 )
= 1.3010
Thus pH of the solution would be 1.3010
Molarity = 0.1M
thus moles of NaOH = 0.10.75 = 0.075mol [using formula mole = C (concentration in M) V in L]
HCl volume = 250 mL or 0.25 L
molarity = 0.5 M
Moles of HCl = 0.50.25 = 0.125 mol
1 mole of NaOH exactly neutralizes 1 mole of HCl according to the reaction
NaOH + HCl NaCl + H2O
Thus 0.075 moles of NaOH will neutralise 0.075 moles of HCl. Therefore HCl is in excess.
Amount of HCl remaining = 0.125 - 0.075 = 0.05 moles.
Thus concentration of H+ ions =
= 0.05 / 1 ( since volume = 0.25 + 0.75 = 1 L)
= 0.05 M
Thus pH= - log[H+]
where [H+] is concentration of H+ ions in molarity
pH = - log (0.05 )
= 1.3010
Thus pH of the solution would be 1.3010