Calculate the pH of each of the following solutions: **just need help solving d

Question

Calculate the pH of each of the following solutions:

**just need help solving d and e**

(a) 0.1000M Propanoic acid( HC3H5O2, Ka= 1.3x10-5)
(b) 0.1000M sodium propanoate (Na C3H5O2)
(c) 0.1000M HC3H5O2 and 0.1000M Na C3H5O2
(d) After 0.020 mol of HCl is added to 1.00 L solution of (a) and (b) above.
(e) After 0.020 mol of NaOH is added to 1.00 L solution of (a) and (b) above.

Answer for d) a. pH=1.70 b. pH=5.49
Answer for e) a. pH=4.29 b. pH=12.30

I need to see the steps of how to solve this!

Explanation / Answer

sol: HONH2 + H2O ? HONH3+ + OH- Before add HCl: 0.400 . . . . . . . . . . 0. . . . .........trace After add HCl: . .0.380 . . . . . . . . . . 0.020 . . . . .trace Kb = [HONH3+]*[OH-]/[HONH2] = (0.020/0.380)*[OH-] =1.1E-8 [OH-] = 0.209E-6... ->...pOH = 6.68 pH+pOH = 14 ...->...pH = 7.32 HONH3Cl is the acid in this conjugated acid-base couple. Hence the added HCl will not react with HONH3Cl and [H3O+] = 0.020. [H3O+] = 0.020. Therefore pH = -log(0.020) = 1.70

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