If 2.50 mL of 1.00 M NaOH are added to 100.0 mL of 100.0 mM aceticacid (pKa = 4.

Question

If 2.50 mL of 1.00 M NaOH are added to 100.0 mL of 100.0 mM aceticacid (pKa = 4.74)
initially buffered at pH 4.00, what would be the final pH of thesolution?

For this question, I started thinking that the HHasselbach equationis needed. Conceptually, I'm not sure why you use that HH equation.But so I find the HA/A ratio to be .18197. That is in 0.1L.
I'm not sure if I need an ICE table?
I'm not sure how to interpret the stoichiometry of thereaction.

I need some help with the CONCEPTS here so I can do these problemsin the future.

Thanks!

Explanation / Answer

No . of milli moles of NaOH n = Molarity * Volume inmL                                                = 1M * 2.5 mL                                                = 2.5 mmol No . of mMoles of CH3COOH , n ' = 100mM * 100 mL                                                        = 100 * 10^-3 M * 100 mL                                                        = 10 mmol
                                        NaOH + CH3COOH ------> CH3COONa + H2O initialconc                      2.5mmol         10mmol                    0                 Equiconc.                           2.5             10-2.5                       2.5                                                                  = 7.5 According to Henderson's Equation pH = pKa + log ([salt] / [acid])                                                             = 4.74 + log ( 2.5 / 7.5 )                                                             = 4.74 -0.48                                                             = 4.26

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