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WINSTONCH [101]
3 years ago
9

What does the Ka represent in the equation and why is it important?

Chemistry
1 answer:
Jet001 [13]3 years ago
5 0

Answer:

The acid dissociation constant, also called the acid-ionization constant, represents the <u>equilibrium constant</u> of a given chemical reaction, which is denoted by the symbol Ka.

It gives the measure of the<u> acidic strength </u>of a chemical substance or the <u>dissociation of an acid</u> in a solution and is represented as the ratio of concentrations of product and the reactants, at equilibrium.

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Diamond cannot conduct electricity but graphite can,why​
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8 0
2 years ago
For the following electrochemical reaction: Al3+(aq) + 3e -&gt; Al(s) Eº = -1.66 V E° = 2.87 F2(g) + 2e -&gt; 2F (aq) Calculate
makvit [3.9K]

<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.

<u>Explanation:</u>

We are given:

E^o_{(F_2/F^-)}=2.87V\\E^o_{(Al^{3+}/Al)}=-1.66V

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.

Aluminium will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

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

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

E^o_{cell}=2.87-(-1.66)=4.53V

Hence, the standard electrode potential of the cell is 4.53 V.

6 0
3 years ago
How many grams of dry nh4cl need to be added to 2.50 l of a 0.800 m solution of ammonia, nh3, to prepare a buffer solution that
tia_tia [17]

Answer:

The correct answer is 574.59 grams.

Explanation:

Based on the given information, the number of moles of NH₃ will be,  

= 2.50 L × 0.800 mol/L

= 2 mol

The given pH of a buffer is 8.53

pH + pOH = 14.00

pOH = 14.00 - pH

pOH = 14.00 - 8.53

pOH = 5.47

The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,  

= -log (1.8 ×10⁻⁵)

= 5.00 - log 1.8

= 5.00 - 0.26

= 4.74

Based on Henderson equation:  

pOH = pKb + log ([salt]/[base])

pOH = pKb + [NH₄⁺]/[NH₃]

5.47 = 4.74 + log ([NH₄⁺]/[NH₃])

log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73

[NH₄⁺]/[NH₃] = 10^0.73= 5.37

[NH₄⁺ = 5.37 × 2 mol = 10.74 mol

Now the mass of dry ammonium chloride required is,  

mass of NH₄Cl = 10.74 mol × 53.5 g/mol

= 574.59 grams.  

8 0
3 years ago
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