A solution that is formed by combining 400 mL of 0.025 molar HCl with 600 mL of 0.015 molar NaOH has a pH of

The reaction: HCl(aq) + NaOH(aq) --> NaCl(aq) + H2O(l)

First you need to know the moles of each constituent added.

moles HCl = Volume(in L) x molarity(in mole/L) = (0.400 L)(0.025 M) = 0.0100 mole HCl
moles NaOH = Volume(in L) x molarity(in mole/L) = (0.600 L)(0.015 M) = 0.00900 mole NaOH

Next you need to determine the moles of excess acid remaining:
0.0100 mole - 0.00900 mole = 0.00100 mole

Now find the new molarity:
molarity = moles HCl/Volume HCl (in L) = (0.00100 mole)/(1.00 L) = 1.00 x 10^-3 M
This will give you the concentration of HCl remaining and [H3O^+]

pH = -log[H3O^+] = -log(1.00 x 10^-3) = 3.00

That's how it's done.

Hope this is helpful to you. JIL HIR

M1 V1 =400X.025= 10/1000 = .01 = [H+]
M2V2 = 600X.015 =9 /1000 =.009 = [ OH-]
.009 moles of H+ ions are neutralised by [OH-] ions of NaOH
SO [H+] =.0 1- .009
= .001= 1x10^-3
pH = - log[10^-3]
= 3 answer

3