A 120.0mL buffer solution is 0.105M in NH3 and 0.135M in NH4Br. What amount of H

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A 120.0 mL buffer solution is 0.105 M in NH3 and 0.135 M in NH4Br, What mass of HCl could this buffer neutralize before the pH fell below 9.00? First calculate the pH of the solution. pH = pKa + log [(B)/(A)] pH = 9.24 + log [.105/0.135] = 9.13. You really don't need that but it's nice to know the starting point. The slip in the pH you have as the minimum (9.00) and calculate base/acid ratio. 9.00 = 9.24 + log [(B)/(A)] I get something like B/A = 0.6 but you need to do it more precisely than that. Then NH3 + H^+ ==> NH4^+ Start with NH3 = 0.105, H^+ = 0 and NH4^+ = 0.135 You want to add x moles H^+ so after the reaction, the quantities are these: NH3 = 0.105-x H^+ = 0 NH4^+ = 0.135+x Substitute these into B/A as follows: (0.105-x)/(0.135+x) = 0.6 and solve for x. Then convert x (in units of molarity) to moles (you had 120 mL solution) then to grams. That should do it. I would check it to make sure by converting g HCl to moles, then to M, then add to NH4 and subtract from NH3 and stick into the HH equation. You should end up with 9.00 or VERY VERY close to it.

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