Question

Worksheet 2 2. Calculate the concentration of each of the following solutions. Determine the molarity (mol/L) of a solution made by diluting 5.880 g of barium chloride to a final volume of 525.0 mL a. b. What is the molarity of a solution that dilutes 12.50 mL of 6.00 M HCI (hydrochloric acid) to a final volume of 1.50 L? If you combine 1.55 g of NH CI (ammonium chloride) with 2.38 g of CaCl, (calcium chlo- ride) and dilute the mixture to a final volume of 721 mL, what is the resulting molarity of chloride ions (mol Cl /L) in the solution? c. 3. A student wants to determine the concentration of an acetic acid (HC,H,O) solution by react- ing it with a 0.450 M NaOH (sodium hydroxide) solution. The student places 25.00 mL of the acetic acid solution in a reaction flask. An indicator is added to the reaction flask. When the reaction goes to completion the solution will change colors because of the indicator. The reaction requires 36.45 mL of the 0.450 M NaOH solution. What is the concentration of the acetic acid solution?

Answer 1

a) Barium Chloride (Bacla) = 5.880gm MM BaCl2 = 137'3 +2(35.45) = 208.2 gm/mol volume = 525 me nBaCl2 = 5880 mol 208.2 Molarity = nBullz x 1000 vim) 5.880 - 208.2 X1000 525 M = 0.054 M Ang 6) Hel » dillution M, = 6M V, = 12:50 me M2 = ? V2 = 1.50 Li isoome Laul Miui= M2UZ 6x12.5 = M2 x 1500 M2 = 0.05M Ane

c) NhuCl = 1.55gm cacl = 2.38gm MMAHUCI = 14+4(1) +35.45 MM CaCl2 = 40.08 +2135.45) =53.49 gm/mol - 110.98 gmi mol nNhuCl = 1.55 mol noaciz = 2.38 mol 53.ug 110.98 ncr = 1.55 mol nei = 2x3.38 53:49 110.98 nei = 0.028gmol ni = 0.061 mol Hotell nci = 0.0289 + 0.061 -) 0.0898 mol volume = 721ml Inci x 1000 0.0898 x 1000 [ci] = Vome) 72) [ci] = 0.125 M/ Ane. inaole] = 36.45 me, 0.45m [CH3COOH] = M, asme nNaOH = (36:45 x 0:45) n(H2COOH = M*25 CH3COOH + NaOH = Cr,coo Na + H2O for completion of Rxh nNaOH = nchacool 36.45 0.45 - Mx25 SM = [CH3COOH) = 0.656 M are