I need assistance with this question please.
"The equation of the line for a graph of voltage (Ecell) versus log[Cu2+] was y= -0.0166x - 0.0082. The voltage of a solution with an unknown [Cu2+] was measured against the same reference cell, under the same conditions; the voltage was found to be 0.0091 V. Determine the [Cu2+] of the unknown solution."
Thanks.
To determine the [Cu2+] of the unknown solution, we can use the equation of the line provided: y = -0.0166x - 0.0082.
Here, y represents the voltage (Ecell) of the unknown solution, and x represents the log[Cu2+].
We are given that the voltage (Ecell) of the unknown solution is 0.0091 V. Plugging this value in for y in the equation, we have:
0.0091 = -0.0166x - 0.0082.
To solve for x, we need to isolate it on one side of the equation. Let's start by moving the -0.0082 term to the other side:
0.0091 + 0.0082 = -0.0166x.
Combining the constants on the left side gives us:
0.0173 = -0.0166x.
To solve for x, we can divide both sides of the equation by -0.0166:
0.0173 / -0.0166 = x.
x ≈ -1.046.
Now that we have the value of x, which represents log[Cu2+], we can solve for [Cu2+] by taking the antilog (inverse log) of x. In this case, we have:
[Cu2+] = 10^(-1.046).
Using a calculator, we find that [Cu2+] ≈ 0.084 M (moles per liter).
Therefore, the approximate [Cu2+] of the unknown solution is 0.084 M.