Question

How many moles of NaOH need to be added to 200.0 mL of a 0.200 M solution of HF to make a buffer with a pH of 2.70? (ka for HF is 6.8 x 10-4) mol 1 | 2 | 3 7 8 9 +/- x 100 Tap here or pull up for additional resources

Answer 1

Please give me better rattings.

Since solution: This is buffer solution made up of weak Anid (HF) and its sult with strong base. pH = 2.70 , Ka = 6.8x104 conc. of HF = 0.2oom Volume 200 ml : 0.22 HF + H+ + F Ka is small neglecting it from And we know, Henderson's equation - - pH = pka + log [ Anim of salt ] o i [Auid] ka=6.8x10 pka = -logka = -log (68x10 or, pka= 3.17 [Acid] - [us] 0.2oom, Salt is NOF, Let a be mole of Noon to be added NaOH + H F Naf + 420 u Using ear o 2-70 = 3017 + log 2012 BRETAGSNIUJE Usin 0:2 08 2.70 = 3.17 + log a or log on you = -0.47 laking Antilog, we get are you = 0.34 h or, 219 0.34*0.4 = 0.136 :. Moles of NaOH to be added is 0.136 moles. WS

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Answer 2

Thank you, but the steps yield an incorrect answer according to the software... automatic feedback says: "The moles of weak acid will decrease when strong base is added."