Calculate the values of pH whe
Calculate the values of pH when 40.0 mL of 0.0250M benzoic acid(HC7H5O2 , Ka=6.3×10-5 ) is titrated with (1) 0.0 mL of 0.050 MNaOH solution? (2) 10.0 mL of 0.050 M NaOH solution? (3) 25.0 mL of0.050 M NaOH solution?
I mostly need help with 2 and 3. The answers are 4.20 and 11.58but I not know how to calculate them.
Answer:
moles of NaOH added= molarity* Volume in L=0.05*10/1000=0.0005
moles of Benzoic acid ( designated as HA). = 0.0250*40/1000=0.001
the reaction between Benzoic acid and NaOH is C6H5COOH +NaOH——->C6H5COONa+ H2O
1 mole of NaOH reacts with 1mole of Benzoic acid. Molar ratio ofC6H5COOH: NaOH= 1:1 ( theoretical)
Actual molar ratio =0.0005:0.001 =1:2
excess is Benzoic acid . All the NaOH gets consumed and moles ofsodium Benzoate ( C6H5COONa) formed =0.0005, moles of Benzoic acidremaining =0.001-0.0005=0.0005
Volume of solution after mixing = 40+10= 50ml= 50/1000L=0.05L
concentrations: Benzoic acid = sodium Benzonate= 0.005/0.05=0.1M
since pH= pKa+ log [ sodium Benzoate/ Benzoic acid]
pKa of benzoic acid =-log (Ka)=4.2
hence pH= 4.2+ log (0.1/0.1)= 4.2
3. when 25ml of 0.05M NaOH is added. moles of NaOH=0.05*25/1000=0.00125, moles of Benzoic acid =0.001
hence excess is NaOH all the Benzoic acid is consumed. Moles ofNaOH remaining =0.00125-0.001= 0.00025
volume of solution after mixing=40+25= 65ml= 65/1000L=0.065L
concentration of NaOH= 0.00025/0.065=0.003846
NaOH being strong base ionizes completely. henceNaOH——>Na++ OH-
[OH-]=0.003845, pOH= 2.41, pH= 14-2.41= 11.59
pOH= -log (0.008)= 2.1
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