Question

30.0mL of 300mM Acetic acid (pKa= 4.76) is mixed with 15.0mL of 210 mM NaOH. What is the pH of the final solution we'll call SOLUTION A?
If 10mL of 0.200M HCl is added to SOLUTION A, what would be the new pH?

Answer 1

Total volume of mixture = 30 + 15 = 45 mL

Molarity of acetic acid in mixture = Initial Molarity * initial volume/ Total volume = 30.0 mL* 300 mM/ 45 mL = 200 mM = 0.200 M

Molarity of NaOH in mixture = 15mL * 210 mM / 45 mL = 70 mM = 0.07 M

	CH3COOH	NaOH	<-->	CH3COONa	H2O
I (M)	0.2	0.07		0	0
C(M)	-0.07	-0.07		+0.07
E(M)	0.13	0		0.07

From Henderson Hasselbalch equation

pH = pK_a + log [CH₃COONa]/[CH₃COOH] = 4.76 + log (0.07/.13) = 4.49

pH of solution A is 4.49

Moles of HCl added = Molarity * Volume = 0.200 M * 10L/1000 = 0.002 moles

HCl is a acid and will react with CH₃COONa to give back CH₃COOH

Moles of CH₃COONa reacted = moles of HCl added = 0.002 mol

Moles of CH₃COONa left = 0.07 - 0.002 = 0.068 moles

Moles of CH₃COOH = 0.13 + 0.002 = 0.132 moles

pH = pK_a + log [CH₃COONa]/[CH₃COOH] = 4.76 + log ( 0.068/0.132) = 4.47 ( New pH)