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Consider a solution containing .2 M HOAc (Ka = 1.8X10^-5) and.10 M NaOAc. Calcul

ID: 679392 • Letter: C

Question

Consider a solution containing .2 M HOAc (Ka = 1.8X10^-5) and.10 M NaOAc. Calculate the pH of this solution and calculate the pHof the solution after .020 mol of HCl (g) is dissolved in 1 L ofthis solution. The acid HOCl has a pKa value of 7.50. Calculate the pH of asolution containing .25 M HOCl and .75 M NaOCl. Here are two problems that I need help with dealing with pHvalues. Thanks. Consider a solution containing .2 M HOAc (Ka = 1.8X10^-5) and.10 M NaOAc. Calculate the pH of this solution and calculate the pHof the solution after .020 mol of HCl (g) is dissolved in 1 L ofthis solution. The acid HOCl has a pKa value of 7.50. Calculate the pH of asolution containing .25 M HOCl and .75 M NaOCl. Here are two problems that I need help with dealing with pHvalues. Thanks.

Explanation / Answer

a )        Formula :                         pH = pKa + log ( No.of mols of salt / No.of mols of acid)    Initial mols of salt   =0.10 mols    Initial mols of acid = 0.20 mols                      pKa   = 4.7447                      pH     = 4.7447 + log ( 0.10 / 0.20 )                                = 4.44 When a strong acid is added to the acidic buffer no.of mols ofsalt decreases and no.of mols of acid increases. No.of mols of salt = 0.10 mols - 0.02 mols                              =0.08 mols No.of mols of acid = 0.2 mols - 0.02 mols                              =0.18 mols                        pH= 4.7447 + log ( 0.08 / 0.18 )                             = 4.39 b) Formula :                         pH = pKa + log [salt]/[acid]                   [salt]    = 0.75 M                   [acid]   = 0.25 M                       pKa = 7.50                       pH   = 7.50 + log ( 0.75 / 0.25 )                               =  7.02