Question

A solution has [c6h5cooh] = 0.100 m and [ca(c6h5coo)2] = 0.200 m. ka = 6.3 × 10−5 for c6h5cooh. the solution volume is 5.00 l. what is the ph of the solution after 10.00 ml of 5.00 m naoh is added?

Answer 1

Answer:

The answer is 4.86

Explanation:

Convert the molarity to moles first.

Moles of C6H5COOH (acid)= 0.100 M x 5 L=0.5 moles of C6H5COO

Moles of Ca(C6H5COO)2= 0.200 M x 5 L= 1 mole of CaC6H5COO (convert this into moles of C6H5COO base by multiplying by the coefficient of two from the molecular formula)

Moles of C6H5COO= 1 mole x 2= 2 moles of C6H5COO

Moles of NaOH= 5.00 M x 0.01 L= 0.05 moles of NaOH

Set up an ICE table using the equation as shown below:

  C6H5COOH         +        NaOH      =        C6H5COO        +    H2O

I:  0.5 moles                     0.05 moles          2.0 moles              N/A (liquid)

C: -0.05 moles               -0.05 moles        +0.05 moles

E:  0.45 moles                 0 moles                2.05 moles

NaOH was the limiting reactant so we subtracted that amount from the C6H5COO and added it to the C6H5COO. A strong base will reduce the concentration of H+ ions from the acid and add OH ions to the concentration of the base.

Then, we use the Henderson - Hasselbalch equation to solve for pH.

pH= pka +log (moles of base/ moles of acid)

pH= -log (6.3 x 10^-5) + log (2.05/0.45)

pH=4.86

Answer 2

6.8 is the pH of the solution after 10 ml of 5M NaOH is added.

Explanation:

Data given:

Molarity of C6H5CCOH = 0.100 M

molarity of ca(c6h5coo)2  = 0.2 M

Ka = 6.3 x 10^-5

first pH is calculated of the buffer solution

pH = pKa+ log 10 $\frac{[A-]}{[HA]}$

pKa = -log10[Ka]

pka = -log[6.3 x10^-5]

pKa = 4.200

putting the values to know pH of the buffer

pH = 4.200 + log 10 $\frac{0.2}{0.1}$

pH = 4.200 + 0.3

    pH  = 4.5 (when NaOH was not added, this is pH of buffer solution)

now the molarity of the solution is calculated after NaOH i.e Mbuffer is added

MbufferVbuffer = Mbase Vbase

putting the values in above equation:

Mbuffer = $\frac{MbaseVbase}{Vbuffer}$

             = $\frac{5X10}{5000}$

             = 0.01 M

molarity or [ A-] = 5M

pH =   pKa+ log 10 $\frac{[A-]}{[HA]}$

pH = 4.200 + log 10 $\frac{5}{0.01}$

pH = 4.200+ 2.69

pH = 6.8