General chem problem?

Question

A solution which is formed by combining 200 mL of 0.15M HCl with 300 mL of 0.09M of NaOH has a [OH-] concentration of? PLease help I absolutely forgot how to do it. Thank you.

Answer 1

total no. of moles of HCl present in 200 ml 0.15 M HCl = 0.15*200/1000 = 0.03 moles.

HCl <==> H(+) + Cl(-)

=> no. of moles of H(+) ions present = no of moles of HCl present = 0.03

similarly , as NaOH ionizes as :

NaOH <==> Na(+) + OH(-)

no of moles of OH(-) present = no of moles of NaOH present = 0.09*300/1000 mole = 0.027 mole

since, H(+) + OH(-) = H2O,
0.03 moles of H(+) will completely neutralise 0.027 moles of OH(-) & (0.03 - 0.027) = 0.003 moles of H(+) will be left.
=>[H(+)] = 0.003/.5
=>pH of the soln = -log[H(+)]
= - log (0.006)
=2.222

since, pH + pOH = 14
here, pOH = 14 - pH = 14 - 2.222 = 11.778
=> - log [OH(-)] = 11.778
=> [OH(-)] = 10^(-11.778) gm ion/L

Answer 2

Convert both of your volumes to liters.

Next calculate the moles of HCl present by multiplying the molarity by the volume (0.15 X 0.2) = 0.03 mol HCl

Next calculate the moles of NaOH used in the same way:
0.09 X 0.3 = 0.027 mol NaOH

Since you have more HCl than NaOH, you will have 0.003 moles of HCl remaining in 0.5 L of solution.

So, [HCl] = 0.003/0.5 = 0.0006 M = [H+]

[OH-] = 1 X 10^-14 / [H+]

(I'm not using a calculator, so you should check all of the calculations.)