Question

A weak acid has a dissociation constant of Ka = 2.50×10⁻².

A. Calculate the degree of dissociation for a 9.7×10⁻² m solution of this acid using the Debye-Huckel limiting law.

B. Calculate the degree of dissociation for a 9.7×10⁻² m solution of this acid that is also 0.200 m in KCl from the Debye-Huckel limiting law using an iterative calculation until the answer is constant in the second decimal place.

C. Repeat the calculation in (B) using the value of γ_± = 0.718 for KCl.

Answer 1

Ans: Given Dissociation constant of weak acid, Kas 215x16". s) ka= 0.025.4 A Given - 2 . of solution. dissociation for 9.9x10 m = 0.097m. Debye-Hucked limiting law, By using Ka(+rt) (m_Y.) m => M' Yg 0.025 - 0.097-m if n=1 then m_ -0.025 =) Ⓡ = (0.47-m)(0-025) -> m= -0.025m + (2425X17) = m 0.025m - (2.425X163)= 0 . -0.025 + 12.0253'-4 ()-(2.425x103)) m 2 ma -0.025 + 0.106 2 m = 0.038 mollkg)

I (2) = 0.03% motika In Yt = -1•113x1 x 10.03% 1 : -0.2295 = 20-2295 =) = 0.1949 when f = 0.1949 them from eq 6 . m” (0-1949) 0.094-m -> m'= 0.039 (0-097-m] = m+0.039m- (3.18x173)=0 m= -0.039 I 10.0393-4(1)(-(3-48x173)) -.0025 ma -0.039 0.128 M: 0.044 mol lkg. I IS (2) = 0.044 nuotkg. In Y = -1-173x1x 10.044 = -0.247 -Y = ē9-247

-> 9 = 0.781 when vi = 018 them from 090 m (0.609) (0-097-m) = m +0.04m - (3:28x10%) = 0 ME -0.04 + 160-045" – 4(1)(-(3-88x163)) -- = 0.025 mo -0.04 + 0.13 2 Im = 0.044 mollig I are obtained. since, same values for 'n I= m (2) = oʻolu molikg. - 0.463 0.044 dissociations, 0.097 OK O ULOU ". = 46-37.

Bikel Z=1 7.31 Given 1:0.2m Activity woefficient is I= 0;2 (2) = 0.2. mellka mY = --173 x 1 x Jo-2 = -0.524 ě0.524 Calculate the conc. of ions produced through dissociation of acid in solution. Therefore from eqo m(0.592,5 0.025 0.097-m : 1m'70.07m - (6:49x103)=0 ma -0.01 + $10.01).-4(1(-(6-49x163) 2 . -0.011 0.179 (M = 0-054 mol kg] M = 0.054 molka

I: 0.2-0.0145 -I = 0-245 mol kg. Y : -1.173X1 X Jo.245 -0.58 -) = 20:58 -) 11 = 0-559 Recalculate 'r' , from M (0.559) eq o = 0.025 0.047-m m+0.08 m - (1-46x103) = 0 m2 -0.08+ (0-085-40176-67:46X163)) 2 m -0.08 +0.143 m= 0.0565 mol/kq | Given the fact that the concentration of kel is only known to a significant figures, this value of m is sufficiently close to the previous value that a further iteration is not necessary. Degree dissociation - 0:0565 x 100 = 58-2% 0.097

S. Given (t = 0.718 from eq m"(0-418) 0.091-m m't 0.04 m - (3088x103) = 0 -0.04 + Jo.04)=-4 (1) 6-(3-88X163)) me - 20.025 i pasiteit ma -0.04 + 0.138 2 Im = 0.049 mol/kg - 1 - 0.049 0:091 :: Degree of dissociation 097 X 100 50-57 30-50%