Question

Be sure to answer all parts. A 35.00-ml solution of 0.2500 MHF is titrated with a standardized 0.1492 M solution of NaOH at 25° C (a) What is the pH of the HF solution before titrant is added? (b) How many milliliters of titrant are required to reach the equivalence point? mL (c) What is the pH at 0.50 mL before the equivalence point? (d) What is the pH at the equivalence point? (e) What is the pH at 0.50 mL after the equivalence point?

Answer 1

wheid sbase . salt HF + NaOH, twa F t to ind imol i mot M, = 0.2500 M2=0:1492 ů, = 35.00m 2 = 9 ka of weak alid HF Å 6.8x104 from tables The Answers may be slightely altered of your ka value is not same.pl remember this. pka__logka @ 1220 =3.17 for weak alids pH = -logjke E log (J6-8x104x0.25 plt --10960.013 ) ñ © At Equivalente point Mill = MzU2 tot 20.8573 ore 2 . Va = 58.6 ml ne

C [HFJust = Mini-M2V2 V = 58.6-0.5 Vitu, -58.1 CHEZ... – 0:25x35-0.1992 35+5801 CHF Icept = 8.758104 [MF] = 0,1492x58.1 -=0.093 3575817 .: pH = pke Hog Isack = 3,17 + log 01093 0875x104 =5.196 (pH = 5-20) At Equivalente point pH = t[ plew + pka floge] @ C = [NA F] = M, V, COD M3.V2 Vitva = 0.25035 = 0.0935 35+58.6 - .: 84-461473-17° +10g 0.0935) -8.07 pH =8.07

♡ pH attent as my atter equivalente pert . Va =5806+0.5=59.1 ml [NadHJexces = M2W 2-0,01 VitV2 .: (NaOH) = 0,1492x59.1-0.25835 59-1 +35 = 7.190 x16 si pot -- log (NaOH) = 3.14 .:PH - 5/14-p04 = 14.00 -3.14 7 pH = 10.84