You have 0.500 l of an 0.250 m acetate buffer solution (i. e. [hc₂h₃o₂] + [c₂h₃o₂⁻] = 0.250 m) at ph 3.50. how many ml of 1.000 m naoh must you add in order to change the ph to 5.25? acetic acid has a pka of 4.74.

Explanation:

Relation between pH and is as follows.

            pH =

Hence, total moles of acetate is calculated as follows.

           = 0.125 M

                 =

                 = 0.0575

Let us assume that there are x moles of acid. Hence, moles of B are as follows.

                 B = (0.125 - x) moles

      = 0.0575

                  x = 0.118 moles

Now, we will calculate the concentration as follows.

           5.25 = 4.74 +

                        = 3.236

                     x =

                         = 0.0295 moles

Therefore, moles of NaOH needed to add are as follows.

        = (0.118 - 0.0295) moles

          = 0.0885 moles

                            =

                            = 88.5 ml

Thus, we can conclude that there is 88.5 ml of 1.000 M NaOH must you add in order to change the pH to 5.25.

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.