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2 years ago

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Caffeine (C_8H_10N_4O_2) is a weak base with a K_b value of 4 times 10^-4. The pH of a 0.01 M solution of caffeine is in the ran

ge of: a. 2-3 b. 5-6 c. 7-8 d. 9-10 e. 11-12

1 answer:

docker41 [41]2 years ago

3 0

Answer:

The pH of the solution lies from 11 to 12.Hence, option e is correct.

Explanation:

The value of $K_b$ for caffine = $4\times 10^{-4}$

$CafOH(aq)\rightleftharpoons Caf(aq)+OH^-(aq)$

Initial

0 0.01 M 0

AT equilibrium:

x (0.01 -x)M x

$K_b=\frac{x(0.01-x)}{(x)}$

$4\times 10^{-4}=\frac{x(0.01-x)}{(x)}$

Solving for x:

x = 0.0096 M

The pOH of the solution is given by :

$pOH=-\log[OH^-}$

$pOH=-\log[x]$

$pOH=-\log[0.0096]$

pOH = 2.02

pH= 14 - pOH = 14 - 2.02 = 11.98

The pH of the solution lies from 11 to 12.

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<u>Answer:</u> The specific heat of metal is 2.34 J/g°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 35 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{35mL}\\\\\text{Mass of water}=(1g/mL\times 35mL)=35g$

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 4.82 g

$m_2$ = mass of water = 35 g

$T_{final}$ = final temperature = 34.5°C

$T_1$ = initial temperature of metal = 115°C

$T_2$ = initial temperature of water = 28.7°C

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$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$4.82\times c_1\times (34.5-110)=-[35\times 4.186\times (34.5-28.7)]$

$c_1=2.34J/g^oC$

Hence, the specific heat of metal is 2.34 J/g°C

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The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

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From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

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There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

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* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

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* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

$K`=~10^{1.12]$

Reaction can take place

2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

$K~=~10^{ -3.56}$

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

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* with OH-

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pK = 19 - 15.74

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