Calculate the pH when 15 mL of 0.26 M HCl is titrated with 45 mL of 0.10 M NaOH.

Question

Calculate the pH when 15 mL of 0.26 M HCl is titrated with 45 mL of 0.10 M NaOH.

Answer 1

HCl & NaOH react together to completeness, no equilibrium of products and both reactants, and consume each other at a 1:1 ratio (as seen by balanced reaction equation).

1) Write balanced reaction

HCl + NaOH --> H₂O + NaCl

2) Determine #moles within particular reaction

15mL * 0.26 mol = 0.0039 moles HCl

1000mL

45mL * 0.10mol = 0.0045 moles NaOH

1000mL

3) Since 1 mole of HCl would react with 1 mole of NaOH, an equal amount of each will be consumed. What remains? Use an ICE table to help (Initial, Change, Equilibrium)

HCl NaOH

I 0.0039 moles 0.0045 moles

C -0.0039 moles -0.0039 moles

E 0 moles 0.0006 moles

0.0006 moles of NaOH will remain.

4) Find new molarity (moles/liter) in total of both liquids:

15mL + 45mL = 60mL = 0.060L ==> 0.0006 moles = 0.01M NaOH = 1 x 10^-2M NaOH

0.060L

5) Turn this number into pH: (remember, pH = -log([H⁺]) )

>> easier way to start -- first find the pOH, because

1. the concentration of OH^- ion is easy to recognize in this reaction
2. pH + pOH = 14

pOH = - log ( [OH-]) ==> pOH = - log(1x10^-2) = 2

pH + 2 = 14 ==> pH = 12