# 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 =  0.1$×$0.75 = 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 = 0.5$×$0.25 = 0.125 mol

1 mole of NaOH exactly neutralizes 1 mole of HCl according to the reaction
NaOH + HCl $\to$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 ​Hions in molarity

pH = - log (0.05 )
= 1.3010

Thus pH of the solution would be 1.3010

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