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vivado [14]
3 years ago
14

Calculate the solubility of ( = ) in moles per liter. Ignore any acid–base properties. s = mol/L Calculate the solubility of ( =

) in moles per liter. Ignore any acid–base properties. s = mol/L Calculate the solubility of ( = ) in moles per liter. Ignore any acid–base properties. s = mol/L
Chemistry
1 answer:
BaLLatris [955]3 years ago
4 0

This is an incomplete question, here is a complete question.

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.

CaCO₃, Ksp = 8.7 × 10⁻⁹

Answer : The solubility of CaCO₃ is, 9.33\times 10^{-5}mol/L

Explanation :

As we know that CaCO₃ dissociates to give Ca^{2+} ion and CO_3^{2-} ion.

The solubility equilibrium reaction will be:

CaCO_3\rightleftharpoons Ca^{2+}+CO_3^{2-}

The expression for solubility constant for this reaction will be,

K_{sp}=[Ca^{2+}][CO_3^{2-}]

Let solubility of CaCO₃ be, 's'

K_{sp}=(s)\times (s)

K_{sp}=s^2

8.7\times 10^{-9}=s^2

s=9.33\times 10^{-5}mol/L

Therefore, the solubility of CaCO₃ is, 9.33\times 10^{-5}mol/L

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Sulfonation of benzene has the following mechanism: (1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3 [fast] (2) SO3 + C6H6 → H(C6H5+)SO3− [slow]
ziro4ka [17]

Question is incomplete, complete question is as follows :

Complete Question : .Sulfonation of benzene has the following mechanism:

(1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3

[fast]

(2) SO3 + C6H6 → H(C6H5+)SO3−

[slow]

(3) H(C6H5+)SO3− + HSO4− → C6H5SO3− + H2SO4

[fast]

(4) C6H5SO3− + H3O+ → C6H5SO3H + H2O

[fast]

write the overall rate law for the initial rate of the reaction as a fraction.

Rate=k(________/_________)

Answer:

The overall rate law for the initial reaction is = k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]

Explanation :

Frist of all, all the common terms are cancelled out and written the overall reaction.

As we know that the rate depednant step is the slowest step of the reaction, rate law is :

                        rate = k_{2} [SO_{3}][C_{6}H_{6}]

But the problem is that SO3 cannot be written in the overall rate law because it is an intermediate.

Rate law for synthesis of S03 is as follows :

                       rate = k_{1}[H_{2}SO_{4}]^{2}

Hence when we substitute equation 2 in equation one,

                   Rate comes out to be =  k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]

6 0
3 years ago
A formic acid buffer solution contains 0. 20 m h c o o h hcooh and 0. 24 m h c o o − hcoox−. the pka of formic acid is 3. 75. wh
irina1246 [14]

A buffer solution contains an equivalent amount of acid and base. The pH of the solution with an acid dissociation constant (pKa) value of 3.75 is 3.82.

<h3>What is pH?</h3>

The amount of hydrogen or the proton ion in the solution is expressed by the pH. It is given by the sum of pKa and the log of the concentration of acid and bases.

Given,

Concentration of salt [HCOO⁻] = 0.24 M

Concentration of acid [HCOOH] = 0.20 M

The acid dissociation constant (pKa) = 3.75

pH is calculated from the Hendersons equation as,

pH = pKa + log [salt] ÷ [acid]

pH = 3.75 + log [0.24] ÷ [0.20]

= 3.75 + log (1.2)

= 3.75 + 0.079

= 3.82

Therefore, 3.82 is the pH of the buffer.

Learn more about pH here:

brainly.com/question/27181245

#SPJ4

6 0
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