Calculate the ph of 1.0 l of the original buffer upon addition of 0.010 mol of solid naoh to the original buffer solution.

This is an incomplete question, here is a complete question.

Consider a buffer solution that is 0.50 M in NH₃ and 0.20 M in NH₄Cl. For ammonia, pKb = 4.75.

Calculate the pH of 1.0 L of the original buffer upon addition of 0.010 mol of solid NaOH to the original buffer solution.

Answer : The pH of solution is, 9.68

Explanation : Given,

Concentration of NH₃ (base) = 0.50 M

Concentration of NH₄Cl (salt) = 0.20 M

Concentration of NaOH =

After addition of NaOH:

Concentration of NH₃ (base) = 0.50 + 0.01 = 0.51 M

Concentration of NH₄Cl (salt) = 0.20 - 0.01 = 0.19 M

First we have to calculate the value of pOH.

Using Henderson Hesselbach equation :

Now put all the given values in this expression, we get:

Now we have to calculate the pH.

Therefore, the pH of solution is, 9.68

idk

explanation:

dasdad

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.