Question

500.0 mL of 0.150 M NaOH is added to 605 mL of 0.200 M weak acid (Ka = 5.15 × 10-5). What is the pH of the resulting buffer?

Answer 1

Answer – We are given, weak acid [HA] = 0.200 M , volume = 605 mL

Ka = 5.15E-5, [NaOH] = 0.150 M , volume of NaOH = 500.0 mL

Calculation of moles –

First we need to convert the given volume to L

Volume of weak acid

We know,

1 mL = 0.001 L

So, 605 mL = ?

= 0.605 L

Volume of NaOH

We know,

1 mL = 0.001 L

So, 500 mL = ?

= 0.500 L

Moles of weak acid HA = molarity of HA x volume of HA(L)

                                        = 0.200 M x 0.605 L

                                       = 0.121 moles

Moles of weak acid NaOH = molarity of NaOH x volume of NaOH(L)

                                        = 0.150 M x 0.500 L

                                       = 0.075 moles

Now we know the reaction between weak acid and NaOH as follow –

HA + NaOH -----> A^- + H₂O

0.121 0.075           0.075

We know limiting reactant is NaOH, since it gives lowest moles for the product

So, moles of weak acid remaining = 0.121 -0.075

                                                        = 0.046moles

Total volume = 605+500

                       = 1105 mL

                       = 1.105 L

New molarity –

[HA] = 0.046 moles / 1.105 L

         = 0.0416 M

[A^-] = 0.075 moles / 1.105 L

        = 0.0679 M

Now we need to use the Henderson Hasselbalch equation-

pH = pKa + log [A^-] /[HA]

calculation of pKa

we know formula

pKa = -log Ka

       = -log 5.15E-5

       = 4.29

Now plugging the values in the formula

pH = 4.29 + log 0.0679 M / 0.0416 M

      = 4.5

So, the pH of the resulting buffer is 4.50