[gracelee]

This is problem that I have been having a really hard time with.  And I need to figure it out before upcoming exam, finals, etc. in case it shows up! Basically- you have a solution containing both Ca²⁺ and Mg²⁺ which you titrate with EDTA at pH 10, from this you find that the combined concentration of [Ca²⁺ + Mg²⁺] = 0.01523M (that part was easy.) Next, you take the same solution, dilute by factor of 20 with KN0₃, and measure the resulting solution with a calcium-ion selective electrode, getting a voltage of -41.4mV. Meanwhile...you have already created a standard curve for the calcium ion selective electrode using standard solutions of CaNO₃. The equation of the best-fit line is (y = 29.6x + 80.4). (fyi- the curve is mV vs. Log[Ca²⁺]). The electrode's selectivity coefficient is K_Ca2+, _Mg2+ = 0.00060. So..we want to find the individual concentrations of [Ca²⁺] and [Mg²⁺] in the original solution.

My thinking-- we have the concentration [Mg²⁺ + Ca²⁺]. My immediate thought is plug the voltage, -41.4mV, into the standard curve equation and determine [Ca²⁺]. Then subtract it from the concentration of [Ca²⁺ + Mg²⁺]. But this obviously doesn't take into account the fact that the measured voltage, -41.4mV is not just a response to Ca²⁺, it's the electrode's response to both Mg²⁺ and Ca²⁺!

Possible methods of solution --> I'm thinking if I could use the selectivity coefficient to determine the error in the voltage due to Magnesium (but I'm not sure how I would do that)- then I could use the error-adjusted voltage to find [Ca²⁺], using the standard curve, and then subtract that value from the combined concentration of Mg²⁺ and Ca²⁺.
OR, I'm wondering if maybe I need to use some kind of system of simultaneous equations?  I could do this but I'm still hung up on the fact that I think the measured voltage, -41.4mV, as it is, shouldn't be used to determine Ca²⁺ from the standard curve because it's not all due to Ca²⁺!
Any help/guidance with this would be greatly appreciated! 

[Borek]

Simultaneous equations is a way to go IMHO. Can you write an equation that shows how the ISE responds to both ions present? It will be a variant of the Nernst equation, just using both concentrations and the selectivity coefficient.

[gracelee]

Thank you for your reply!

Okay, so equations would be:

E = Constant + (0.05916/2)Log([Ca²⁺] + K_{Ca2+, Mg2+}[Mg²⁺])

and

[Mg²⁺ + Ca²⁺] = 0.01523

I kept getting hung up on the standard curve...but maybe the purpose of it is just so the constant can be calculated?