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Annette [7]
3 years ago
11

A buffer solution contains 0.368 M hydrocyanic acid and 0.360 M potassium cyanide . If 0.0513 moles of sodium hydroxide are adde

d to 225 mL of this buffer, what is the pH of the resulting solution
Chemistry
1 answer:
MrRa [10]3 years ago
7 0

Solution :

Millimoles of hydrocyanic acid = 225 x 0.368

                                                   = 82.8

Millimoles of potassium cyanide = 225 x 0.360

                                                   = 81

Millimoles of sodium hydroxide = 51.3

Therefore,

pOH = pKb + log [salt - C / bas + C]

        = 4.74 + log[82.8 - 51.3 / 81 + 51.3]

         = 4.102

Therefore, pH = 9.05

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Mixture with two or more substances that are evenly mixed
sertanlavr [38]
A homogeneous mixture, the substances are uniformly distributed throughout the mixture
8 0
3 years ago
Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant
antoniya [11.8K]

Answer: The value of E^{o} for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

          AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)

Its solubility product will be as follows.

       K_{sp} = [Ag^{+}][Cl^{-}]

Cell reaction for this equation is as follows.

     Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)

Reduction half-reaction: Ag^{+} + 1e^{-} \rightarrow Ag(s),  E^{o}_{Ag^{+}/Ag} = 0.799 V

Oxidation half-reaction: Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-},   E^{o}_{AgCl/Ag} = ?

Cell reaction: Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product, K_{sp} = [Ag^{+}][Cl^{-}]

                                               = 1.77 \times 10^{-10}

Therefore, according to the Nernst equation

           E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

At equilibrium, E_{cell} = 0.00 V

Putting the given values into the above formula as follows.

         E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

        0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}    

       E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}

                  = 0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

           E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}

       0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}

       0.577 V = 0.799 V - E^{o}_{AgCl/Ag}

       E^{o}_{AgCl/Ag} = 0.222 V

Thus, we can conclude that value of E^{o} for the half-cell reaction is 0.222 V.

3 0
3 years ago
A sample of benzene was vaporized at 25◦C. When 37.5 kJ of heat was supplied, 95.0 g of the liquid benzene vaporized. What is th
grandymaker [24]

Answer:

Enthalpy of vaporization = 30.8 kj/mol

Explanation:

Given data:

Mass of benzene = 95.0 g

Heat evolved = 37.5 KJ

Enthalpy of vaporization = ?

Solution:

Molar mass of benzene = 78 g/mol

Number of moles = mass/ molar mass

Number of moles = 95 g/ 78 g/mol

Number of moles = 1.218 mol

Enthalpy of vaporization =  37.5 KJ/1.218 mol

Enthalpy of vaporization = 30.8 kj/mol

8 0
3 years ago
Which type of emission has the highest penetrating power?
Savatey [412]
You may recall that gamma radiation is at the far right (high energy) end of the electromagnetic spectrum. Because of its high energy level, gamma radiation has the greatest penetrative power.
7 0
3 years ago
An elephant eats about 500 pounds of grass and leaves every day. there are 2 african elephants and 3 asian elephants living in t
denpristay [2]
Lets think about this for a moment
500lb per 2 african and 3 asian elepahnts so 500 * 5 > 2500LB
That is 1 day but 7 days are in a week so this is what we need > simple > 2500 * 7 = 17,500lbs a week.
6 0
3 years ago
Read 2 more answers
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