What is the ionic strength of 0.2 M Na2HPO4 in a solution?
7 answers
https://en.wikipedia.org/wiki/Ionic_strength
0.8
The ionic strength of a 0.2 M Na2HPO4, solution will
be
be
Ionic strength of 0.2 M Na2HPO4 is 0.6 M
0.6 M is the ionic strength.
0.2 M * 2 *(1)^2+ 0.2 M * 1 * (-1)^2
= 0.6
0.2 M * 2 *(1)^2+ 0.2 M * 1 * (-1)^2
= 0.6
=1/2 [0.2x2x1*2 + 0.2x1x(-2)*2]
=1/2 [0.4 + 0.8]
=1/2 [1.2]
= 0.6
=1/2 [0.4 + 0.8]
=1/2 [1.2]
= 0.6
I=1/2{C1Z1^2+C2Z2^2+......}
=1/2{0.2×2×1^2+0.2×1×(-1)^2+0.2×1×(-3)^2
=1/2{0.4+0.2+1.8}
=1/2{2.4}
=1.2M
=1/2{0.2×2×1^2+0.2×1×(-1)^2+0.2×1×(-3)^2
=1/2{0.4+0.2+1.8}
=1/2{2.4}
=1.2M