Question 6. 1M CH3COOH solution. The dissociation constant of acetic acid is 1.8 x 10 -5 . Answer: pH = – log[H + ] For weak acids,
Question 7. Find the pH of a buffer solution containing 0.20 mole per litre sodium acetate and 0.18 mole per litre acetic acid. Ka for acetic acid is 1.8 x 10 -5
Given that Ka = 1.8 x 10 -5 pKa = – log (1.8 x 10 -5 ) = 5 – log 1.8 = 5 – 0.26 = cuatro.74 pH = 4.74 + log \(\frac < 0.20>< 0.18>\) = 4.74 + log \(\frac < 10>< 9>\) = 4.74 + log 10 – log 9 = 4.74 + 1 – 0.95 = 5.74 – 0.95 = 4.79
Question 8. What is the pH regarding an aqueous provider gotten by the fusion 6 gram off acetic acidic and you may 8.2 gram out-of sodium acetate and you will deciding to make the regularity comparable to 500 ml. (Given: K to have acetic acid is 8 x ten) Answer: According to Henderson – Hessalbalch picture,
- this new hydrolysis constant,
- standard of hydrolysis and you may
- pH of O.1 M CH3COONa solution (pKa for CH3COOH is 4.74).
3COONa is a salt of weak acid (CH3COOH) and a strong base (NaOH). Hence, the solutions is alkaline due to hydrolysis. CH3COO (aq) + H2O (aq) \(\rightleftharpoons\) CH3COOH (aq) + OH – (aq) Give that pKa = 4.74 pKa = – log Ka i.e., Ka = antilog of ( – PKa) = antilog of ( – 4.74) = antilog of( – 5 + 0.26) 10 -5 x 1.8
- BaSO4
- Ag2(CrO4)
Determine the new pH of 0
Question 1. Which one of the following buffer is present in blood? (a) CH3COOH + CH3COONa (b) NH4OH + NH4Cl (c) H2CO3 + NaHCO3 (d) HCI + NaCl Answer: (c) H2CO3 + NaHCO3