How many mL of 0.1M HCl should be added to 100mL of a 0.2Msolution of tris base

Question

How many mL of 0.1M HCl should be added to 100mL of a 0.2Msolution of tris base to make a buffer with pH 7.8. What will thefinal tris concentration be? I know tris pKa=8.08 and HCl pKa=8.3. I tried the equationpH=pKa + log (base/acid), but I am not getting it. How many mL of 0.1M HCl should be added to 100mL of a 0.2Msolution of tris base to make a buffer with pH 7.8. What will thefinal tris concentration be? I know tris pKa=8.08 and HCl pKa=8.3. I tried the equationpH=pKa + log (base/acid), but I am not getting it.

Explanation / Answer

Chemical equation :                            HCl + Tris base --------- > salt +H2O From the equation 1 mol of acid reacts with one mole of baseto form 1 mol of salt . No.of mols of acid = x L * 0.1 M                              = 0.1x mols No.of mols of tris base = 0.1 L *0.2 M                                     =0.02 mols No.of mols of salt formed will be equal to the no.of mols ofHCl as it is strong one . No.of mols of salt   = 0.1x mols     No.of mols of base = 0.02 mols -0.1x mols Being a basic buffer , pOH = pKb + log ( No.of mols of salt /No.of mols of base) pH    = 7.8 pOH = 14 - 7.8         = 6.2 pKa = 8.08 pKb  = 14 - 8.08          = 5.92 Upon substituing in the above formula ,    6.2 = 5.92 + log ( 0.1x / 0.02mols - 0.1x ) log ( 0.1x / 0.02 mols - 0.1x ) = 0.28           0.1x /0.02 mols - 0.1x  = 1.9                           0.1x             =0.038 - 0.19 x                                    0.29x = 0.038                                            x = 0.038 / 0.29                                               = 0.131 L                                               = 131mL                  

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