If 10 ml of 0.5 M NaOH is added to 10 ml of 0.5 Macetic acid, what is the pH of

Question

If 10 ml of 0.5 M NaOH is added to 10 ml of 0.5 Macetic acid, what is the pH of the solution? pKa=4.76 The answer is 9.08 can you please explain the equation fromanalyzing the acid base relationship to the equation to use and thesteps to get the answer, thank you If 10 ml of 0.5 M NaOH is added to 10 ml of 0.5 Macetic acid, what is the pH of the solution? pKa=4.76 The answer is 9.08 can you please explain the equation fromanalyzing the acid base relationship to the equation to use and thesteps to get the answer, thank you

Explanation / Answer

If 10 ml of 0.5 M NaOH is added to 10 ml of 0.5 M aceticacid, we know that molarity is given by; no of moles / totalvolume now no of moles of NaOH = total volume taken * molarity                                         = 0.01 L * 0.5 M                                           =5 x 10-3 moles now no of moles of acetic acid = total volume taken* molarity                                         = 0.01 L * 0.5 M                                           =5 x 10-3 moles so both react to give same 5 x 10-3 moles ofsolution but now total volume is 20 ml = 0.02 L so now molarity of solution = 5 x 10-3 /0.02                                         = 0.25 M so now as salt is made up of weak acid and strong base pH after hydolysis can be calculated as: pH = 1/2( pKw + pKa + log10C) where, pKa = -log Ka        = 4.76 pKw = 14 ( its value for water) and C is concentration of salt in moles/ litre         here C = 0.25M ( as calculated) so pH = 1/2( 4.76 + 14 + log10( 0.25))           =9.08 so  pH of the solution = 9.08 now no of moles of acetic acid = total volume taken* molarity                                         = 0.01 L * 0.5 M                                           =5 x 10-3 moles so both react to give same 5 x 10-3 moles ofsolution but now total volume is 20 ml = 0.02 L so now molarity of solution = 5 x 10-3 /0.02                                         = 0.25 M so now as salt is made up of weak acid and strong base pH after hydolysis can be calculated as: pH = 1/2( pKw + pKa + log10C) where, pKa = -log Ka        = 4.76 pKw = 14 ( its value for water) and C is concentration of salt in moles/ litre         here C = 0.25M ( as calculated) so pH = 1/2( 4.76 + 14 + log10( 0.25))           =9.08 so  pH of the solution = 9.08 pH after hydolysis can be calculated as: pH = 1/2( pKw + pKa + log10C) where, pKa = -log Ka        = 4.76 pKw = 14 ( its value for water) and C is concentration of salt in moles/ litre         here C = 0.25M ( as calculated) so pH = 1/2( 4.76 + 14 + log10( 0.25))           =9.08 so  pH of the solution = 9.08 so pH = 1/2( 4.76 + 14 + log10( 0.25))           =9.08 so  pH of the solution = 9.08

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