the equation of the line for a graph of voltage versus log[Cu2+] was y= - 0.0179x - 0.0015. 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 0.012 V. Determine the [ Cu2+] of the unknown solution.
To determine the [Cu2+] of the unknown solution, we need to use the equation of the line and the given voltage measurement.
The equation given is:
y = -0.0179x - 0.0015
where y represents the voltage and x represents log[Cu2+].
We know that the voltage of the unknown solution is 0.012 V. So we can plug this value into the equation and solve for x.
0.012 = -0.0179x - 0.0015
To isolate x, we can move the constant term to the other side:
0.012 + 0.0015 = -0.0179x
0.0135 = -0.0179x
To solve for x, divide both sides by -0.0179:
x = 0.0135 / -0.0179
x ≈ -0.75307
Now we have the value of x, which represents log[Cu2+]. To find the [Cu2+], we need to convert log[Cu2+] back into a regular number.
Using the logarithmic relationship, we can rewrite the equation as:
log[Cu2+] = x
[Cu2+] = 10^x
Now, substitute the value of x into the equation:
[Cu2+] = 10^(-0.75307)
[Cu2+] ≈ 0.190 μM (micromolar)
Therefore, the [Cu2+] of the unknown solution is approximately 0.190 μM.
To determine the [Cu2+] of the unknown solution, we can use the equation of the line for the graph of voltage versus log[Cu2+]. In this case, we have the equation:
y = -0.0179x - 0.0015
Where:
y is the voltage of the solution with the unknown [Cu2+]
x is the log[Cu2+]
We are given that the voltage (y) of the unknown solution is 0.012 V. Let's substitute this value into the equation:
0.012 = -0.0179x - 0.0015
To solve for x, we need to isolate x on one side of the equation. Let's start by moving the constant term to the other side:
0.012 + 0.0015 = -0.0179x
0.0135 = -0.0179x
Now, divide both sides by -0.0179 to solve for x:
x = 0.0135 / -0.0179
x ≈ -0.753
The value of x represents log[Cu2+]. To determine the [Cu2+], we need to convert x back to its original form using the logarithmic function. Since the logarithmic function used was base 10, we can use the antilog function or use the inverse logarithm with base 10 (10^x).
[ Cu2+ ] = 10^x
Now, substitute the value of x into the equation:
[ Cu2+ ] ≈ 10^(-0.753)
Using a calculator, we find that:
[ Cu2+ ] ≈ 0.186 M
Therefore, the [ Cu2+ ] of the unknown solution is approximately 0.186 M.