Quiz 5 Total 20 points For calculation questions, show equations ...

Quiz 5 Total 20 pointsFor calculation questions, show equations and calculations. We will give partial points even if youranswer is incorrect. For multiple choice, circle one answer.I 7 points (a: 4 points, b 3 points)Answer the following questions. K w is a dissociation constant of water (1.0 × 10 -14 ).(a) There is a solution of 2.0 × 10 -8 M NaOH (Note: this is a very dilute solution). Choose acorrect relationship that represents [H + ] of this solution from (i-v). Briefly explain why therelationship is correct. x = [H + ].(i) x = K w /2.0 × 10 -8 (ii) Log(x) = (K w /2.0 × 10 -8 ) (iii) (x + 2.0 × 10 -8 )x = K w (iv)2.0×10 -8 x =K w(iv) None of the aboveReasoningBecause solution is very dilute, dissociation of H 2 O is not negligible(or because 2.0 × 10 8 < 1.0×10 7 =(K w ) 1/2 )(b) There is a solution of 0.20 M of ammonia. pK a of the ammonium ion is 9.24 (not pK b ).Choose the most appropriate formula that represents [OH - ] of this solution.(i) (0.20 × 9.24) 1/2 (ii) (0.20 K w /10 -9.24 ) 1/2 (iii) (0.20 × 10 -9.24 ) 1/2 (iv) (0.20 × 10 9.24 ) 1/2(iv) 0.20 × 10 -9.24(iii) (FKa) 1/2 = (0.20 10 -9.24 ) 1 pointII 6 points (a: 2 points, b: 4 points)We titrate 10.0 mL of 0.100 M dibasic base (B) with 0.100 M HCl. pK b1 = 4.3 and pK b2 = 8.0.Suppose V a mL of the HCl solution is titrated.(a) Choose the value closest to the pH obtained when V a = 10.0 mL.(i) 5.0 (ii) 5.5 (iii) 6.0 (iv) 6.5 (v) 7.0 (vi) 7.5 pH ~ (pK 1 + pK 2 )/2 =7.85pK 1 = 14.0 – pK b2 = 6.0pK 2 = 14.0 – pK b1 = 9.7pH ~ (pKb 1 + pKb 2 )/2 =6.15 (1 point)(b) Calculate the pH when V a = 2.0 mLpH = pK a2 + Log[B]/[BH]. (1 points)K a2 = K w /K b1 pK a2 = pK w – pK b1 = 14.0 - 4.3 = 9.7 (1 point)[B]/[BH] = (Ve – Va)/Va = (10.0 -2.0)/2.0 = 4.0 (1 point) pH = 9.7 + Log4.0 = 10.3 (1point)

Your Name . Your section (Circle one)M/W AM, M/W PM, T/R AM, T/R PMIII 7 points (a: 3 points, b: 4 points)Calculate [Cu 2+ ] at the following points in the titration of 50.0 mL of 0.040 M Cu(NO 3 ) 2 (bufferedat pH 5.00) with V mL of 0.20 M EDTA, where K f = 10 18.80 and α Y4- =3.7×10 -7 . (a) V = 5.0 mLand (b) V =10.0 mL(a) V e = 50.0 mL × 0.040 M /0.20 M = 10 mL So this is the case before equivalence point ( 1point)[Cu 2+ ] = {(Ve – V)/Ve} (0.040 M) {50.0 mL/55.0 mL } (1 point)= 0.018 M(b) This is the equivalence point.[CuY 2- ] = 0.040 M {50.0 mL /60.0 mL} (1 point)= 0.033 MCu 2+ + EDTA CuY2-x x 0.033 –x[CuY 2- ]/{[Cu 2+ ][EDTA]} = K f ’ (1 point)(0.033 –x)/x 2 = K f ’K f ’ = 3.7×10 -7 ×10 18.80 = 3.7× 10 11.8 = 2.3 × 10 12(1 point)x = (0.033/ K f ’) 1/2 M = 1.2 × 10 -7 M (1 point)