Question

Calculate the concentration of all species in a 0.14 M KF solution.
Ka(HF)=6.3×10−4

Express your answer using two significant figures. Enter your answers numerically separated by commas.

Calculate the concentration of all species in a 0.14 MKF solution. K. (HF) = 6.3 x 10 4 Express your answer using two significant figures. Enter your answers numerically separated by commas. PO ALV x x 8 O a ? x x-10 x x x = K+], F-), HF), [OH], H30+] = M Your submission doesn't have the correct number of answers. Answers should be separated with a comma. No credit lost. Try again.

Answer 1

The question is solved using simple concept of ionic equilibrium.

Solution - The values given in question are O . Concentration of K = 014 M ka CMF) 2. 6.3X10-4 We know that 11 kaxkhz 10? kez 1014 to 12 ka ndoll 6.3X107 31.58 X10 0.44M KE contains o.19 mkt fo.14ME F + H2068) = HE HOH 0 0 initialD equilibrium 0 -14 0.14-21 [MF] [on] $ 120) can be ignored [F] [42010) xxx kez 1o. 14-2) 1.5881014 x² ai (0.14-2) 2022 0.19 X 1.88 X10" - 1:58 X 10" 11 S 8X10 - 2-21 X 10 20

deviding through out the equation by 1:58x10" we get 6-3284667? 4 2 - 0.14 zo ☺ The equation @ is like ax²+60146 zo which can be salved as az -6 ± √ 6²-qac ca Hence equation ☺ can be solved as 4 -1 V1-4 (6-3281d")x (0-14) 2XL 3XToto 22-1+VI+ 353 XL010 1-26 X 10 11 3.53X1010 >>> 1 chemce 1+3:53X100 3:53 X1010 - + V3.33 X 10 x 2 1.26 XToll -1 + 1-87 X10S 1:26X toll alt 1.87 XOS 22 1-26 X1011 to sign is not taken because ton ecubration t is not negative 4-87X.10% >> E-1) June -1 +1-87x10x 1.87 xlos

xz 1687x105 1-26 X 10" 1:48X166 405 X 15 CM 22 [on] = [HEJ z 1:5X100M Now [F] z 0.19 - 1.58 10 6 0.14 >>7 1 58106 Head (FJn 0.14 M. We know that [hot] ton] z 10 [4g 04 J = 10 10 18 a ToHJ l'SX106 04 HITUISTITIHIHIHIHNUTII [M30*]z-667x16M Hence the answer is - [kt],[EL[HE],[04] , [M30*) = 0.14M, 0-14M, 1:5X10CM, 1:5x16CM 6.678100M