Question

Using the table given below for Ka values, compare the pH of an HF buffer that contains 0.13 MHF and 5.0x10-2 M NaF with another HF buffer that contains 1.0x10-2 MHF and 0.16 M NaF. Ka and K b values for selected weak acids and bases HF Acid Ka Phosphoric acid H3PO4 7.5 x 10-3 Hydrofluoric acid 7.2 x 10-4 Nitrous acid HNO2 4.5 x 10-4 Formic acid HCHO 1.8 x 10-4 Acetic acid HC2H3O2 1.8 10-5 Carbonic acid H2CO3 4.3 x 10-7 Hydrosulfuric acid H2S 9.1 x 10-8 Dihydrogen phosphate H2PO4 - 6.2 x 10-8 Hydrocyanic acid HCN 4.9 x 10-10 Hydrogen phosphate HPO42 2.2 x 10-13 Base |Kb Methylamine CH3 – NH2 4.4 x 10-4 Carbonate CO3 2.2 x 10-4 Ammonia 1.8 x 10-5 c acid NH:

What is the pH of a buffer solution containing 0.13 M HF and 5.0×10⁻² M NaF?

What is the pH of a buffer solution containing 1.0×10⁻² M HF and 0.16 M NaF?

Answer 1

Hi, dear student please find your answer if you have any query feel free to ask.

ANSWER- As we know, to calculate the pH. and poh values of Buffer solution, we use the Hander som equation – Handerson equation for weak Acid and salt of strong Base- [salt ] pH = pka + log 10 AidT and pka = - logioka In the given question ka= 7.2x 10-t for HF Pka = – logio6.2x10-4] = -log10 7.2 + 4 logio Pka = -0.85+ 4 = 3.15 HA) <D> 0.13 M HF and 5.0X10-2M Naf [HE] = 0.13, [NaF] = 5:0X10? Prane) = 3.15 on putting these values in Handersm equation - P.H = 3.15+ log10 (3.0x10 10.13 ] pH = 3.15 + log105 - 2 logo - log, [0.13] = 3:15 +0.6989-2 +0.88 = 4.7289- 2.0000 pH= 2:7289 on Rounding of B- pH= 273

> Fox 10-2M HE and 0:16M NaF HF] = l'ox10-2 [NaF] = 0.16M on putting these values in Handersm equation- (salt pH = Pka + log10 Acid pH = 3.15 + logioch.ox 70-27 PH=3.15+ log [0.16] - log, [ox lo-2 pH=3.15+ logio [016] - logio +2 logiolo pH = 3.15+ log 0.16 - 0+2 pH=3.15-0.79+2 pH = 5.15-0.79 pH= 4.36 pH = 4.36