What is the final pH if .2 mol HCL is added to .5L of a buffer solution formulat

Question

What is the final pH if .2 mol HCL is added to .5L of a buffer solution formulated using .28 M NH3 and .22 M NH4CL? Assume there is no change in volume upon addition of HCL. Kb(NH3)= 1.8e-5

Explanation / Answer

THIS WILL HELP YOU Initial Concentration of the Buffer Solution: Now NaClO dissolves to completion so we have an initial concentration of ClO: 1 M * .4 L [ClO]_initial: ------------- = .4 M (.6 + .4)L .6 M * .6 L [HClO]_initial: ------------- = .36 M (.6 + .4)L HClO H+ + ClO- ---- -- ---- .36 0 .4 initial -x x x change .36-x x .4+x equilibrium x(.4+x) so Ka = --------- ==> x = 1.206 X 10^-5 (.36-x) which is [H+] (its concentration) and for ClO-, its new concentration is .400012 in 1 Liter, thus there are .4 moles of ClO now. After HCl is added: Moles of HCl added: 3.00M * .04 L = .12 moles These .12 moles will react with the .4 moles of ClO and will be used up so there will be no change in the number of moles of H+. There is, however, a change in concentration. For if we have a concentration of 1.206 X 10^-5 M for H+, then our new concentration is: -5 1.206 X 10 * 1 L -5 [H+] = --------------------- = 1.15954 X 10 (1 + .04) L pH = -log([H+]) = 4.936

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