Question

Indicate the concentration of each ion present in the solution formed by mixing the following.

Part A

44.0 mL of 0.110 M Na2SO4 and 25.0 mL of 0.200 M KCl. Assume that the volumes are additive. Enter your answers, separated by commas, using three significant figures.

Part B

3.80 g KCl in 75.0 mL of 0.230 M CaCl2 solution. Assume that the volumes are additive. Enter your answers, separated by commas, using three significant figures.

Answer 1

2 — >kt kt tu caa) Post A & molarity = no. of moles (mol) volume of solution (L) = M = - = = MXV no. of moles = mass (9) molar mass (g/mol) in Molarity = mass of colute 9 molarmars of solute (g/mol) volume of solution (2) Na, SA cara Natag + 504 cag. M = 0.110 14 = 0.110 mol/L ; v = 44.0mL = 0.044 L was so 4 = MNasso X V Nassou = (0.110 mol/L) (0.044 4) = 0.00464 mol The ratio of Na, soy: wat = 102 so not = 2(0.00484ml) and ratio of = 0.00968 mof Nascoy: 504 = 1.1 → so n Sor2 = 0.00964 mos Ки Cag) M - 0.200 M = 0.20 mol/L; V=25.0mL = 0.025L ker = (0,20 mol/L) Co.or5h) = 0.005 mol Mku: Mkt ner = 1:1:1 . Mkt = 0.005 mol no = 0.005 mol Now calculate the molarity of all rous, total volume is 44 mL +25mL – 600mt = 0.069t # Maut = 0.00968 mo. 0.069 L = 0.140 mol/L 0.00484 mol 0.069 L = 0.070 mol/L Mkt - 0.005 mol = 0.069 2 Mi = 0.005 mol = 0.072 mol/L 0.069 L Wat Mso2 = 0.072 mol/L

= 3.89 74.559/mol = 0.051 mot I Part B 3.89 of kl. ku - Kto & cu car) nka= mass of ku molar mass of kel - - 74.5g/mol = 0 tau : Mat once = lotil net = 0.051 mot na = 0.051 mol and calls a cat + 20 car) M-0.230m=0.23 mol/L : v= 75.0mL = 0.075L Marely = (0.23 mol/L) (0.075L) = 0.01725 mol # Mcache & Cat2 = 18 so nat2 = 0.01725 mol neachy & na = 182, so Me - = 2(0.01725 ind) = 0.0345 mol Now volume of solution - 75 mL = 0.075L = 0.051 mol = 0.68 mol/L 0.075 [cat2] = 0.01725 mol = 0.23 mollL 0.075 L 0.0855 mil the rurl= 0.051molt 0.0345 mol = 0.075 L 0.0751 -
for part A, 3 significant figures

[Na+] = 0.14 mol/L

[SO4 2-] = 0.07 mol/L

[K+]= 0.07 mol/L

[Cl-]= 0.07 mol/L.

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