A 32.7mL sample of HCl of an unknown concentration wasdiluted to 100mL. A 20.0mL
ID: 685717 • Letter: A
Question
A 32.7mL sample of HCl of an unknown concentration wasdiluted to 100mL. A 20.0mL sample of the diluted solution wastitrated against 0.879M solution of Ba(OH)2. Thetitration required 15.2 mL. Calculate the molarity of the initialsolution. I got the diluted solution concentration to be1.39M but i am really lost on where to go from here. How do Icalculate the molarity of the initial solution? A 32.7mL sample of HCl of an unknown concentration wasdiluted to 100mL. A 20.0mL sample of the diluted solution wastitrated against 0.879M solution of Ba(OH)2. Thetitration required 15.2 mL. Calculate the molarity of the initialsolution. I got the diluted solution concentration to be1.39M but i am really lost on where to go from here. How do Icalculate the molarity of the initial solution?Explanation / Answer
Let the unknow concentration of the HCl = x Therefore the concentration of the HCl after dilution =32.7* x mol / 100 mL = 0.327 x M we take the 20 mL of the solution and titrated with the0.879 M Ba(OH)2., Hence - Number of moles of HCl consumed in the reaction = 0.327x mol/L *0.020 L = 0.00654 x mol Since the balanced equation of the titration - Ba(OH)2 (Aq) + 2 HCl (aq) --------->BaCl2 (aq) + 2 H2O (l) In this titration, the number of moles of bariumchloride consumed = 0.0152 L * 0.879 mol/L = 0.0133608 mol According to given equation, we need 0.01336 mol*2 =0.02672 mol Therefore - 0.00654 x = 0.02672 Thus , x = 0.02672 /0.00654 =4.08 M Hence the concentration of the HCl = 4.0 MRelated Questions
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