Question

A volume of 6.0 mLs of 4.0 N NaOH (FW = 40g/mol) is added to a solution containing 2.50g of tri-protonated Lysine Acid (FW = 147 g/mol). What is the pH of the final solution?

Answer 1

Milliequivalence of NaOH = Normality * volume = 4 * 6 = 24

since Normality = (weight/Equivalent)1000/volume in ml weight

normality * volume = (weight /eqivalent weight)1000 = milliequivalence

Milliequivalence of Tri-protonated lysine = (weight / equivalent weight)1000

equivalent weight = molecular weight/no. of replaceable hydrogens for an acid = 147/3 = 49

                                                                = (2.5/49)*1000

                                                                = 51.02.04

here the limiting reagent is NaOH so it will be consumed completely when added to 2.5g of triprotonated lysine, so the remaining lysine will show acidic pH.

Milliequivalence of lysine left = 51.0204 - 24 = 27.0204

here if know the volume of lysine solution then we can calculate the exact pH of final solution

I assume volume of lysine as V ml then total volume of solution after mixing will be (V+6)ml

Normality of lysine = 27.0204/(V+6)

Molarity of [H⁺] = Normality* 3 = 81.0612/(V+6)

pH = -log[H⁺]

substitute [H⁺] in the above equation you will get pH of final solution