10ML of .010M HF solution titrated with a .010M solution of NAOH at 25Celcius. K

Question

10ML of .010M HF solution titrated with a .010M solution of NAOH at 25Celcius. Ka HF =6.8x10^-4. 1.) How many mL of the base is required to reach the equivalence point? 2) what is the pH of the HF solution before base is added? 3) What is the pH at eqivalence point? 4) what is the pH at 1/2 volume to equivalence point?

Explanation / Answer

HF(aq) --> H+(aq) + F-(aq) Ka = 6.8 * 10^-4 [HF] initial = 0.10 mole/L = x moles/ 0.010 L ==> x moles = 0.001 moles HF a) x ml * 0.01 moles NaOH/1000ml = 0.001 moles H+ x = 100 ml of NaOH is required. Ans. b) before the base was added, pH = - log [F-] = - log (0.01) = 2.0 Ans. c) mole NaOH left = 0.01 - 0.001 moles = 0.009 moles HF moles OH- generated = 0.001 moles final volume = 10 ml + 100 ml = 50 ml = 0.110 L [NaOH] final = 0.009 moles NaOH/0.100 L = 0.09 M [OH-] final = 0.001 moles OH-/0.100 L = 0.01 M pH = pKa + log[OH-]/[NaOH] = 3.16 + log(0.01/0.09) = 2.20 Ans.

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