Asked Oct 22, 2019

A 25.0ml sample of a 0.3000 M solution of aqueous trimethylamine is titrated with a 0.3750 M solution of HCl. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pKb of (CH3)3N = 4.19 at 25oC.

1. pH after 10.0mL of acid have been added

2. pH after 20.0mL of acid have been added

3. pH after 30.0mL of acid have been added


Expert Answer

Step 1

pH on adding 10.0mL HCl is calculated as follows,


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Initialmoles N= /3 Concentration of CH Total volume 0.300M 25mL (25+10)mL 0.21M 0.3750M 10 mL (25+10)mL Concentration of HCl=Initialmoles Total volume - 0.107M (сн), (CH) NH HCl Cl + /3 Initial 0.21M 0.107M 0 0 Change Equilibrium 0.21M-x -X -X X х 0.107M-x 0.107M 0.107M Since, HCl is limiting reagent, x0.107M (CH) NH CHN ВН = pKb +log РОН-рКъ +log BJ 0.107M 4.19 0.0165 4.2 4.19log 0.103 pH 14-pOH 14-4.2 рH -9.8

Step 2

pH on adding 20.0mL HCl is c...


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Initial moles N- Total volme Concentration of CH1 0.300Mx25mL 0.167M (25+20 mL Initia moles 03750M 20mL Total volume - 0.167M Concentrationof HC1 (25+20 mL (сн - HCl CH NH CI Initial Change Equilibrium 0.16 7M- 0.167M 0.167M 0 1 0.167Mx 0.167M 0.167M NH pH will be due to CH, NH H,O CH3 Cн. N + Initial 0.167M 0 0 Change Equilibrium 0.16 7M- pk pli-pk -14-4.19-9.81 K-10-981-155x10-10 N Kg (CH NH assumingx'tobevery sm all 155x10-10, 0.167Mx 2-155x10-10 0.167M)-0.25885 x10-10 x-v0.25 885x10-10-0.509x10-5 pHo,0]-log (0.50910-5 -5.29 pH-5.29


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