A selective electrode for Ca+2 is submerged in a solution where [Ca+2] = 6mM. The meter reads 65.3mV. A reading is made in a solution which has been diluted on a 1/25 ratio from the original solution you want to brown townyze and gives off a reading of 0,0432V.
The electrode responds to the following ecuation: E = L - (59.1mV/n) * pCa+2
Calculate [Ca+2]
The answer is 810.15mM. I got 0.437mM
I did the following:
E = L - (59.1 * 65.3mV)/2e- * 2.22pCa+2
E = L - 4283.74 mV*pCa+2/e-
L - 4283.74 mV*pCa+2/e- = L - (59.1*43.2mV)/2e- * pCa+2
4283.74 = 1276.56 * pCa+2
pCa+2 = 3.36
[Ca+2] = 4.37e-4 M
[Ca+2] = 0,437mM
I've found a few errors as I was typing this but I am unsure how to actually solve the problem. I was absent the class my teacher explained the electrobrown townitic methods.
