Question
Indicate the concentration of each ion present in the solutionformed by mixing: 1) 42.0mL of .140 M NaOH and 37.6mL of .360M NaOH M Na+, M OH- 2)44.0mL of .140 M Na2SO4 and25.0 mL of .200 M KCl M Na+, MSO42-, M K+, MCl- 3)3.60 g KCl in 75.0 mL of 0.270 M CaCl2solution. Assume that the volumes are additive. M K+, M Ca2+,M Cl- Indicate the concentration of each ion present in the solutionformed by mixing: 1) 42.0mL of .140 M NaOH and 37.6mL of .360M NaOH M Na+, M OH- 2)44.0mL of .140 M Na2SO4 and25.0 mL of .200 M KCl M Na+, MSO42-, M K+, MCl- 3)3.60 g KCl in 75.0 mL of 0.270 M CaCl2solution. Assume that the volumes are additive. M K+, M Ca2+,M Cl- 3)3.60 g KCl in 75.0 mL of 0.270 M CaCl2solution. Assume that the volumes are additive. M K+, M Ca2+,M Cl-
Explanation / Answer
1) The number of moles of NaOH in thesolution = 0.042 L*0.140 mol/L + 0.0376 L * 0.360mol/L = 0.00588 mol + 0.013536 mol = 0.019416 mol Total volume = 42 mL + 37.6 mL = 79.6 mL = 0.0796 L Therefore numbe of moles of Na+ / Cl- in the solution = 0.019416 mol / 0.0796 L = 0.2439M 2) Number of moles ofNa2SO4 = 0.044 L*0.140 mol/L = 0.00616 mol Numberof moles of KCl = 0.025*0.2 = 0.05 mol Therefore - [Na+] = 2*0.00616 / (0.044+0.025) = 0.1785 M [SO4-2] = 0.00616 mol / 0.069 L = 0.08927 M [K+]= [Cl-] = 0.05/0.069 = 0.07246 M 3) Number of moles of KCl = 3.60 g / 74.45 g/mol = 0.04835 mol Number of moles of CaCl2 = 0.270 mol/L * 0.075 L = 0.02025 mol Therfore [K + ] = 0.04835 mol/0.075L = 0.646 M [Cl-] = 2*0.02025 mol + 0.04835 mol /0.075L = 1.184 M [Ca+2] =0.270M