CHEM162L-Homework 8- measuring the pH during titration This problem set will hel

Question

CHEM162L-Homework 8- measuring the pH during titration This problem set will help you understand the three different scenarios that can happen during a (Weak Acid + Strong Base) titration to help you answer the last question on the post lab. I. In a beaker, there is 40 mL of 0.1M weak acid, HA (Ka 2.5 x 10). Calculate the p of this acid! Then, calculate the pH after (a) 25 mL. (b) 40 mL and (c) 45 mL of O.IM NaOH have been added Original pH of the weak acid: HAH+A C: E: Ka-9.5/o solve x = 5. pH-4,30 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 1. What is the total volume? Hint: Vol of HA+NaOH. Use MIVI M2V2 Step 2, Use the total vol and calculate new concentrations [OH]- [HA] = Step 3. Express the neutralization reaction HA +OH' - A. +11,0 what species present? Notice this 1" scenario: at equilibrium both [AJ and [HA] are present, hence we can use Henderson-Hasselbach to find pH! Step 4. Calculate pH based on the scenario

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

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