Question

1. A buffer is 0.100 M in HF and 0.100 M in NaF. When a small amount of nitric acid is added the pH only slightly drops. Write the chemical equation that shows the added nitric acid being neutralized by this buffer.

2. What is the pH of a buffer that is 0.120 M formic acid (HCHO2) and 0.080 M in potassium formate (KCHO2)? The Ka of formic acid is 1.8 x 10^ -4 .

3. The curve shows the titration of a weak acid with a strong base. Assume that the acid is monoprotic and the base is “mono”hydroxide. Answer the following: (a) What is the pH at equivalence? (b) What is the pKa of the weak acid? (c) Which part of the curve is it necessary to use Kb to calculate pH (d) Which part of the curve would it be good to use the Henderson-Hasselbalch equation to calculate pH? Circle it on the plot

pH 0 20 40 60 80 Volume of base added (mL)

4. A 500 mL buffer solution is 0.10 M of a weak acid HA and 0.10 M of A - and has an initial pH = 4.19. What is the pH of the buffer upon addition of 0.010 mol of NaOH?

Answer 1

HF as Ans ²-1. Our buffer contains hydrofluoric acid HE weak acid and sodium fluouide NaF salt of its Conjugate pase and fluoride anion of - H Frage H Pagt F raq When nitric acid is added assuming no change eu volune a buffer this and will be neutralised by NaF, the chemical equation is as follows. - HNO₂ + NaF Ž HF + NaNO₃ . This will increase HFF concentration and deen NaF concentration. Anso- 2. our buffer contain's and potassium formate formic acid, THCHO₂) = 0.1200 [KCPO₂] = 0.080u. HCHO, (ag) ² CHO2 (aq) + H t cag) - Ka = 1.8810-4 Ka = [CHO 2 ] [H+] [HC HU₂] 1.8 x 10-4 = 0.080) CH+] (0.120) [H+] = 1.8 x 104 & 0.120 (0.08) [H+] = 2.7 x 10-4 pH = -log[ht] = -log (2.74004). pH = 4 log (2.7) pH = 3.568

This Ans: -3 -3 This is curve for titration of weak and strong acid base, with base is added to solution Curitially only weak base) Tas At equivalance point all weak and is neutralized and get converted to its conjugate base. However it pH at equivalence point is not equal to 7. This is due to production of conjugate base during titration, resulting solution is slightly basic. The endpoint and equivalance point are not exactly saine: the equivalance pount u determaued by staichometry of reaction. At equivalance pH & 8 (grater than 7 due to basic sol".) (b) pka for weak acid is determined by pH at half equivalance point. As pH at half equivalance point = pka. Half equiralance point is at half volume of base as compored to equivalence porut ! Here it is difficult to analyse out, - pka. po pH at half equi, point ñ 4.5. = so pka ñ 4.5. (o since at equilibrium we find solution be to be basic so to find pH we must consider reaction of base formed from week acid and how to find ko and pH of solution at "equilibrium and onwards!

Date: 1 1 Page no (a by Henderson - Hassel balch equation, - pH = pka + log ( THAT CAI A where HA is This is applicable Concentration to acid and when beeth be zero. is its aniou KAY and CHAT fart of " This equation is vale valed after start o titrapt to equivalance point Because before addition of titraut nearly zero concentration of auiou eu case of a meak arid at and at equilibrium all acid gets neutralised .

4. Volume = soonl. [HA= 0.lom, CA = 0.014 initially, pH = 4.19 By thenderson- Hasselbalch equation pH = pka + log ((A-7 1 THAI pka = pH + log ( [A]) = 4,19 +log (0.1) pka=4,19 to= 4.19 Now, moles of HA initially = 0.1ox 500 x103 = 0.05 molt - moles of À initrally = 0.10 x 300 x103 = 0.05 molt NaOH added will react with HA in 1 1 ratio and decrease it's moles and adde increase some no. of moles of base A. - NaOH + HoH → Not + H₂O + A I 0.000 0.05 . o 0.05 E o 0.05-0.010 0.05 +0.010 moles of HA= 0.05 -0.010 = 0.04 mol. moles o A = 0.05 +0.010 = 0.06 mol

CHAJO.Olax x103 = 0.08 m. SOD [A] = 0.06 x 103 = 0.12 M. soo Again, pH = pka & log / [A3/CHAJ) pH = 4, 19 t log ( 0 12/0.08) DHE 4.19 + 0.176 pH = 4.366