Question

Hello I need help with number # 2. as it’s asking me to make predictions and fill in the first column of Part A of this experiment

d. 2.5 xIO 10.t 2. Make predictions and fill in the first column of Part A of this experiment 3. What is the app roximate expected pH ofa 0.030 M HNo, solution? hat is the final pH after 1 drop (0.05 mL) of 6 M HCl is added to 1.01 freshly prepared pure wat a significant pH change? 0.05ml (demı) ( er that was originally at a pH of 7.0. Is there ): .000 3 mol H

Please help as I don’t know what problem two means by predicting for the first column?

Answer 1

2).ans:- We know that, pH of a solution is the Negative logarithm of hydronium ion or hydrogen ion concentration.i.e.,

pH = -log [H₃O⁺]or -log [H⁺]

0.1 M HCl = 10^-1 M HCl ------> H⁺ + Cl^-

here, [H⁺] = 10^-1 M , Therefore, pH = -log [10^-1] = - (-1) = 1

0.01 M HCl =  10^-2 M HCl ------> H⁺ + Cl^-

here ,[H⁺] = 10^-2 M Therefore, pH = -log [10^-2] = - (-2) = 2

0.001 M HCl = 10^-3 M HCl ------> H⁺ + Cl^-

here ,[H⁺] = 10^-3 M Therefore, pH = -log [10^-3] = - (-3) = 3

0.1 M HC₂H₃O₂or  10^--1 M CH₃COOH(Acetic acid)   ------> CH₃COO^- + H⁺

Acetic acid is weak acid therefore , It does not dissociate completely in aqueous solution and thus, pH is not 1 as for 0.1 M HCl. In such cases for weak acids or bases Ostwald’s Dilution Law is used .i.e.,

pH = - log[H⁺] = -log Cα and where, α=√ (K_a / C)

Where , C= concentration, K_a = Dissociation Consant , α = Degree of dissociation

K_a for acetic Acid is given as = 1.8 x 10^-5

Thus,   α =  √ (K_a / C) = √ 1.8 x 10^-5 / 0.1 = 1.34 x 10^-2

pH = -log ( 0.1 X 1.34 x 10^-2 ) = 2.87

Thus, pH of 0.1 M of HC₂H₃O₂ is 2.87.

0.1 M NH₄Cl It is a salt made from NH₃ and HCl. It is thus, a salt of weak base and strong acid. pH for such salt is given by the following equation,

pH = 7 - 1/2 pK_b -1/2 log C

pK_b for NH₄Cl is = 4.74 (pK_b = -log K_b) and C = 0.1 M

Putting the values in above equation and pH obtained for 0.1 M NH₄Cl is = 5.13

pH of Deionised water = Neutral i.e., 7

pH of Tap water = it is generally less than the neutral i.e. less than the 7 . or we can say somewhere between 6 and 7.

pH of 0.1 M NaCl = It is a neutral salt as it is formed of srong acid HCl and strong base NaOH. THus pH of 0.1 M NaCl is 7.

pH of 0.1 M NaC₂H₃O₂= It is a salt of weak acid CH₃COOH and strong Base NaOH.pH for this salt is given as,

pH = 7 + 1/2 pKa + 1/2 log C

Thus, pH = 5.13

pH of 0.1 M NaHCO₃ = 8.3

pH of 0.1 M Na₂CO_{3 =} It is an basic salt hence pH is 11.6

pH of 0.1 M NH₃ = It is moderate Base = approximately 12

pH of 0.1 M NaOH = 13 as NaOH------> Na⁺ + OH^- .it is a strong base and produce OH^- ions in the solution .

By knowing the values of K_a and K_b we can easily predict the pH values.