Question

determine the comcentration for all solutions.

Solution: Potential: 102M 82 mV 10-'M 134 mV 104M 190 mV 10M 203 m V 10%M 243 mV 10-7M 262 mV Unknown #5 176 mV

Answer 1

Following is the - complete Answer -&- Explanation: for the given: Question: in...typed format...

$\Rightarrow$Answer:'

Concentration of Unknown:( when potential = 176 mV ) = 0.000591562 M  ( mol/L )

i.e. as the table shown below, under Explanation, and under : Answer.  below...

$\Rightarrow$Explanation:

Following is the complete Explanation: for the above Answer.

• Given:

Following is the given: data set, in the form of a table: prepared by using MS Excel, presented in ...image format...

Number of Solutions Concentration of Solution, M (mol/L) Potential, mV (milli-volt) 1.00E-01 82 1.00E-02 134 1.00E-03 1901 1.00E-04 203 1.00E-05 243 1.00E-06 2621

$\Rightarrow$Note: The above table, gives the values of the potential ( in milli-volt, mV ) and the corresponding, concentrations: of the first solution: to the seventh solution, as given: in...image format...

$\Rightarrow$Now, we will have to make a calibration curve, i.e. make a graph: of Potential ( mV ) vs. Concentration ( mol/L ) of the given solutions, to determine the : linear equation of the TrendLine, and then use the Equation: to find the value of the : Concentration: of the 3rd solution: with given Potential ( 176 mV ) ...

• Step - 1:

Using the above, table, we will have to take logaritms: of the given concentrations: to modify the above table: as required:

Number of Solutions 82 Concentration of Solution, M (mol/L) Potential, mV (milli-volt) log [ concentration] 1.00E-01 1.00E-02 134 1.00E-03 1901 1.00E-04 203 1.00E-05 243 1.00E-06 262
​​​​​​​

• Step - 2:

​​​​​​​Now, if we plot: the potential ( mV ) vs. log₁₀ [conc. ] , based on the above table; we will get the following graph, prepared using MS Excel, presented here in......image format...

Potential ( mV) vs. log (conc.) plot 300 250 200 Potential / mV y = -35.42x + 61.66 1 50 + — Potential (mv) vs.log (conc.) plot Linear (Potential (mv) vs. log (conc.) plot) / TE 0 log10(conc.)

$\Rightarrow$ Linear equation of : TrendLine:   is... y = -35.42 x + 61.66 -----------------Equation - 1

$\Rightarrow$Where:

1. y =  potential [=] mV , milli-volt
2. x =  Logarithm ( base 10 ) , of the, given: concentrations , in M ( mol/L )

• Step - 3:

Now, if we plug-in: y = 176, in the above : Equation - 1, we will get the following value:

$\Rightarrow$i.e. x =  Log₁₀ (conc.) =  - 3.228 , i.e. when the potential = 176 mV

• Step - 4:

If we take the anti-log: of the above value of ' x ', we would get the following, concentration:

$\Rightarrow$  i.e. anti-log₁₀ ( x ) = 10^-3.228 =  5.9156 x 10^-4 M ( mol/L ) = concentration...

• Answer:

​​​​​​​Therefore, the following table: prepared by using MS Excel, presented in...image format...will be our Answer...

Number of Solutions 82 نانا | لا ما | يا دانية Concentration of Solution, M (mol/L) Potential, mV (milli-volt) log [ concentration] 1.00E-01 1.00E-02 134 1.00E-03 1901 1.00E-04 203 1.00E-05 243 1.00E-06 262 0.000591562 176) -3.228

$\Rightarrow$As we have found above: the concentration, when the value of : potential = 176 mV , will be the following:

$\Rightarrow$Concentration of Unknown:( when potential = 176 mV ) = 0.000591562 M  ( mol/L )