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3 years ago

If I add 475mL of water to 75.0mL of a 0.315M NaOH solution, what will the molarity of the diluted solution be?

1 answer:

5 0

Answer is: the molarity of the diluted solution 0,043 M.
V(NaOH) = 75 mL ÷ 1000 mL/L = 0,075 L.
c(NaOH) = 0,315 M = 0,315 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0,075 L · 0,315 mol/L.
n(NaOH) = 0,023625 mol.
V(solution) = 0,475 L + 0,75 L.
c(solution) = 0,023625 mol ÷ 0,550 L.
c(solution) = 0,043 mol/L.

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Use the density to convert volume into mass.

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2 years ago

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Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

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The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

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To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

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3 years ago

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