Question

We dissolve 2.77 g of an unknown acid, HA, in enough water to produce 25.0 mL of solution. The pH of this solution of HA(aq) is 1.33. We titrate this solution with a 0.250 M solution of NaOH. It takes 43.7 mL of the NaOH solution to reach the equivalence point. (a) (2 points) What is the molar mass of HA? (b) (2 points) What is the pK, value of HA(aq)? (c) (2 points) What is the pH at the equivalence point? (d) (2 point) What is the pH one third along the titration when you still have twice as much HA(aq) as A (aq)?

Answer 1

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El.997475388 X 10 Xo 15902 45 45 a) moles of molar mass b) molarity of HA = NaOH used moles of HA Present = 43. 7x10 3x0,250 (mol) of HA= 2.77 9 = 0.010925 mol - 253.5469108 g/mol = 254 glmal 0.010925 mol 0.010925 mol 26.0x10 3 L = 0.437 M PH = -dog (tz ot] =-dog (kon . c) skal = 10- (10-133) ² Ka = 5.0063195 06 *103 -PH =10/33 0.437 ³ pka = -log ka =-dog (5.006319506x103) = 2.300481437 ~ 2.30 Kp 10-14 1014 Ka 5.0063195064 10 3 -12 = 1.997455388 x 10 [A] = 0.010925 mod て O. 159024545 m [OH (25.0+43-7)x1032 k, x = 99 7+3R x 20 -12 = 5.6360271*10^9 m pOH = - dog low") -- 10gf5.6360291807) = 6.2490 26927 PH = 14 - Pot = 14-6.249026925 - J. 750973073 ~ 7.75 d) Using HH equation, PH = Pkat log[A] [HA 2.300481437 + 109 = 2.300481437 - 0301029995 = 1.999451442 ~ 2.00 2