A. The measured pH of a 0.100 M HCl solution at 25 degrees Celsius is 1.092. From this information, calculate the activity coefficient of H+.B. The measured pH of a solution of 0.010 HCl and 0.090 KCl at 25 degree Celsius is 2.102. Calculate the activity coefficient of H+ in this solution. C. Why does the pH change in part B relative to that in part A?

activity coefficient

activity coefficient

The change in pH in part A = 0.092

The change in pH in part B =  0.102

Explanation:

From the given information:

pH of HCl solution = 1.092

Activity of the pH solution [a] =

[a] = 0.0809 M

Recall that [a] = × C

where;

 = activity coefficient

C = concentration

Making the activity coefficient the subject of the formula, we have:

B.

The pH of a solution of HCl and KCl = 2.102

[a] =

[a] = 0.00791 M

activity coefficient

C. The change in pH in part A = 1.091 - 1.0 = 0.092

The change in pH in part B = 2.102 -2.00 = 0.102

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g