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Charra [1.4K]
3 years ago
8

A buffer solution is made using a weak acid, HA. If the pH of the buffer is 9 and the ratio of A– to HA is 100, what is the pKa

of HA?
Chemistry
1 answer:
sdas [7]3 years ago
3 0
To calculate the pKa of the weak acid, we use the Henderson-Hasselbalch equation. It is expressed as pH = pKa - log [HA]/[A-]. This equation takes into account the concentration of the substance that does not dissociates into ions since it is a weak acid. We caculate as follows:

pH = pKa - log [HA]/[A-]
9 = pKa - log 1/100
pKa = 7
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What is the mass of 5 mole of ammonia . Calculate the number of NH₃ molecules, nitrogen atom and hydrogen atoms in it..
Aloiza [94]

Molar mass of NH_3

\\ \sf\longmapsto 14u+3(1u)

\\ \sf\longmapsto 14u+3u

\\ \sf\longmapsto 17g/mol

We know.

No of moles=Given mass/Molar mass

\\ \sf\longmapsto Given\;Mass=17(5)

\\ \sf\longmapsto Given \:Mass\:of\:NH_3=85g

Now

Lets write the balanced equation

\\ \sf\longmapsto N_2+3H_2=2NH_3

  • There is 2moles of Ammonia
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Now

\boxed{\sf No\:of\:Molecules =No\:of\;moles\times Avagadro\:no}

For Hydrogen

\\ \sf\longmapsto 3\times 6.023\times 10^{23}

\\ \sf\longmapsto 18.069\times 10^{23}

\\ \sf\longmapsto 1.8\times 10^{22}molecules

For Ammonia

\\ \sf\longmapsto 2\times 6.023\times 10^{23}

\\ \sf\longmapsto 12.046\times 10^{23}

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3 years ago
the normal boiling point of ethanol is 78.3 and its molar heat of vaporization is 393. Calculate the vapor pressure of ethanol a
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<u>0.12 atm</u><u> </u><u>vapor pressure</u><u> of ethanol at 45.0 C.</u>

What is vapor pressure in science definition?

  • Vapour pressure is a measure of the tendency of a material to change into the gaseous or vapour state, and it increases with temperature.
  • The temperature at which the vapour pressure at the surface of a liquid becomes equal to the pressure exerted by the surroundings is called the boiling point of the liquid.

We will use the Clausius-Clapeyron equation,

ln(P2/P1) = dHvap/R[1/T1-1/T2]

where,

P1 = unknown

P2 = 1 atm

T1 = 30 oC = 30 + 273 = 303 K

T2 = 78.3 oC = 78.3 + 273 = 351.3 K

dHvap = 39.3 kJ/mol = 39300 J/mol

R = 8.314 J/K.mol

Feed values,

ln(1/P1) = 39300/8.314[1/303 - 1/351.3]

P1 = 0.12 atm

thus, the vapor pressure at 30° C is 0.12 atm.

Learn more about vapor pressure

brainly.com/question/2510654

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2 years ago
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Answer:

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Explanation:

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Hope this helps

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