Question

help with the following please show all steps, i want to verify my mistake i made. thanks

Question 5 (1 point) Given the following calculations find the ionic strength of a solution containing 0.2165 M MgSO4, 0.001032 M HCl, and 0.1204 M trisodium citrate. Enter your answer in M (mol/L). Your Answer: Answer units

Answer 1

The complete step wise calculation of the ionic strength of the solution is given as below:-

③ It is given that, concentration af mg 50 y = 0.2165m mg soust Mg 2+ (aq) + 504²(ar) Ionic strength can be defined as:- I=lci zi? - o where, I = ionic strength (i= concentration and zi = charge on the ion.

Here, concentration of [mg]ht = (504] 2- -0.2165m charge on mg&t = +2 and charge on sou2-=-2 Then IF I [0.2165 (+2)² + 0.2165 (-2)}] on I, - * *8*0,2165 = UX0-2165 on I1 = 01866) — @

For 0.001032 mnal charge on ht = +1 and c ê=- concentration af [n] = [ce] =0,001032m Iq= + (0.001032(+1)2+0.001032 (-1)2) Then = 0.00 1032 X 2 X 0,001032 on I = 0.001032 — ☺

Finally we have o.1204 m tisodium citrate. Trisodium citrate= Naz C6H5 07 Naz C6H507 368 + C6H5O73- Concentration af Col 5073- = 0.1204 m concentration of 3 Na = 3x 0.1204m Then Iz = + [3x 0.1204 C+1)2 + 0.1204 (-3)) or I3 = _ x0.1204 [1x12] = 6X0.1204 = 0.7224 om (I3 = 0.7224 - 0 So, the Total I = I1+I2+I3 = (0-866)+ (0.001032)+0.7224 or I=1.59 So, the total ionic strength of the solution will be I = 1:59

Since, ionic strength depends upon the concentration of the solution. So, the unit of the ionic strength will be the unit of the concentration. Since, unit of the concentration is M or mol/L. So, the unit of the ionic strength will be M or mol/L. Then the ionic strength of the solution will be; I = 1.59 M = 1.59 mol/L.