If 38.5 mL of 8.9 ✕ 10-3 M HBr is added to 27.0 mL of 4.6 ✕ 10-3 M LiOH, what is the pH of the solution?
4 answers
I got pH = 2.48 ... Please show some effort in solving problem.
I get a pH of 3.66
The pH is 2.47. The above didn’t convert back to total molarity of the acid.
38.5-ml(8.9E-3M HBr) + 27.0-ml(4.6E-3M LiOH) => ?pH of final mix
=> 0.0385(8.9E-3)mole HBr + 0.027(4.6E-3)mole LiOH
=> (3.43E-4) mole HBr + (1.24E-4) mole LiOH
=> (2.188E-4) mole HBr (in excess)* + (1.24E-4) mole LiBr + (1.24E-4) mole H₂O
*LiOH is Limiting Reactant in the HBr + LiOH => LiBr + H₂O Rxn leaving HBr in excess.
[HBr] = 2.188E-4 mole HBr(aq)/(38.5+27.0)ml = (2.188E-4 mole HBr)/(0.0655-L soln) = 0.00334M HBr
LiBr ions do not hydrolyze in aqueous solution, therefore, the final mix pH is due to the 0.00334 M HBr that delivers 0.00334 M H⁺ ions as a strong acid ionizing 100%.
=> pH(final mix) = -log[H⁺] = -log(0.00334) = 2.476
=> 0.0385(8.9E-3)mole HBr + 0.027(4.6E-3)mole LiOH
=> (3.43E-4) mole HBr + (1.24E-4) mole LiOH
=> (2.188E-4) mole HBr (in excess)* + (1.24E-4) mole LiBr + (1.24E-4) mole H₂O
*LiOH is Limiting Reactant in the HBr + LiOH => LiBr + H₂O Rxn leaving HBr in excess.
[HBr] = 2.188E-4 mole HBr(aq)/(38.5+27.0)ml = (2.188E-4 mole HBr)/(0.0655-L soln) = 0.00334M HBr
LiBr ions do not hydrolyze in aqueous solution, therefore, the final mix pH is due to the 0.00334 M HBr that delivers 0.00334 M H⁺ ions as a strong acid ionizing 100%.
=> pH(final mix) = -log[H⁺] = -log(0.00334) = 2.476