CHEM16 This problem Acid + 62L.- Homework 8- measuring the pH during titration S
ID: 546529 • Letter: C
Question
CHEM16 This problem Acid + 62L.- Homework 8- measuring the pH during titration Strong Base) titration to help you answer the last question on the post lab. set will help you understand the three different scenarios that can happen during a (Weak In a beaker, there is 40 mL of 0.1M weak acid, HA (Ka 2.5x 10*). Calculate the pH of this acid! Then, calculate the pH after (a) 25 ml. (b) 40 mL and (c) 45 mL of 0.1M NaOH have been added. Original pH of the weak acid: C: E: Ka = solve x pH = Comments: The following NaOH additions will demonstrate 3 scenarios that can happen at equilbrium. PAY ATTENTION to these scenarios! (a). 25 mL of 0.1 M NaOH added Step I. What is the total volume? Hint: Vol of HA+NaOH. Use MIVI-M2V2 Step 2, Use the total vol and calculate new concentrations [OH]- HAJ Step 3, Express the neutralization reaction HA +OHA+H2O what species present? Notice this 1t scenario: at equilibrium both IA'] and [HA] are present, hence, we can use Henderson-Hasselbach to find pH! Step 4. Calculate pH based on the scenario pH =Explanation / Answer
1)
concentration of acid = 0.10 M
Ka = 2.5 x 10^-8
HA --------------> H+ + A-
0.10 0 0 --------> I
- x x x -----------> C
0.10 - x x x -----------> E
Ka = x^2 / 0.10 - x
2.5 x 10^-8 = x^2 / 0.10 - x
x = 5.01 x 10^-5
[H+] = 5.01 x 10^-5 M
pH = -log (5.01 x 10^-5 )
pH = 4.30
a)
millimoles of acid = 40 x 0.1 = 4
millimoles of base = 25 x 0.1 = 2.5
HA + NaOH ----------> NaA + H2O
4 2.5 0 0
1.5 0 2.5
pH = pKa + log [salt / acid]
= 7.60 + log [2.5 / 1.5]
pH = 7.82
b) 40 mL NaOH :
millimoles of base = 40 x 0.1 = 4
HA + NaOH ----------> NaA + H2O
4 4 0 0
0 0 4
here salt remains. this is equivalence point.
salt concentration = 4 / 40 + 40 = 0.05 M
pH = 7 + 1/2 (pKa + log C)
= 7 + 1/2 (7.60 + log 0.05)
pH = 10.15
d)
millimoles of base = 45 x 0.1 = 4.5
strong base NaOH remains = 0.5
concentration = 0.5 / 40 + 45 = 5.88 x 10^-3 M
pOH = -log (5.88 x 10^-3) = 2.23
pH = 11.77