Answer :
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 [tex]\frac{[A-]}{[HA]}[/tex]
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 [tex]\frac{0.2}{0.1}[/tex]
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 = [tex]\frac{MbaseVbase}{Vbuffer}[/tex]
= [tex]\frac{5X10}{5000}[/tex]
= 0.01 M
molarity or [ A-] = 5M
pH = pKa+ log 10 [tex]\frac{[A-]}{[HA]}[/tex]
pH = 4.200 + log 10 [tex]\frac{5}{0.01}[/tex]
pH = 4.200+ 2.69
pH = 6.8
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