Question

Hydrofluoric acid, HF is a weak acid with K. = 3.5X10. It dissociates according to HFMH *( + Frog (a.) What is the pH of a buffer solution which is 0.30M in F and 0.60M in HF? (b.) What will be the pH of the resulting solution when 2.0mL of 0.5M HCl are added to 50ml of this (from part a) buffer solution? (c.) What will be the pH of the resulting solution when 2.0mL of 0.5M NaOH are added to 50mL of this (from part a) buffer solution?

Answer 1

Sole Given data, Hydrofluoric aad, HA 95 a weak acid with Ka = 3.5*10* HF (aq) = htCag) + F(az) Pka = -logka = -log (3.5x109) = 3.5 Pka = 35 a What is the pH of a buffer Solution 0.30M in F and 0.60M in HF. which is F = 0.30 M HF 20.60M To, calculate the pH We will use Henderson Hassenbach equation p" Pka + log [conjugate base] (acid] Pka 2.3.5 Conjugate base] - F -0.30M (acid] =HF = 0.60m

pH 3.5+ (0.5) 2.3.5 + 60.3 010) (: log (0.5) = log1/2) = log1-1992= 0-loga = -0.3010 log (0-5) -- 0.3010) – 3.5-0.3 by what will be the pH of the resulting Solution When 2.0mL of 0. 5M td are added to 50 ml of this buffer Solution. Moles of F = molarity x volume = C0.30m) x (50 ml) Mole of F = 15M mol. Total volume Mates of HF - 60.60m (50m 50+2 250 mL 2 30m mol Moles of Hal= (0.5m) (ame) 2 AM mol He will give Ht, which will Consume F to give tf.

New moles of F = 15M mol - 1 M mol 214 mmol. F ₂ Concentration moles of F 14 m mol 52mL Total volume 0.27 New moles of HF z 30m molt im mol 231 m mol. Concentration of HF = 31 mmol 50 mL Again 20.60m using Henderson - Hasselbach equation. * - 5 4 23 ( 2 ) 23.5+ log (0.45) * 3,5+(-0.34678749) ? 3.15321 (² log (0.45) value = -0.34 678749) what will be the pH of the resulting 2.0mL of 0.5m wooh are added to 50mL solution this of buffer Solution,

Moles of Naot - (0.5M) x (a.omu) . 1 mmol. Naoth will consume ff and increase F New moles of E 15 mmol + im mo) 216 mmol. New Concentration of F. 16 mmol 5a.mL 20-31m. New moles of HF z 30mmol - Immol 2 ga mol New Concentration of HF = gam mol 5am L. 20.56 m. Again Using Henderson - Hasselbach equation pk 35+10g Cover ? 3.5+ log (0.55351) ? 3.5+ (-0.25963731) C log 0.55 Value 2 -0.26) *** 3.24 2 3.5 -0-26