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il63 [147K]
3 years ago
10

A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3

10−5. What mass of KC6H5CO2 should the student dissolve in the HC6H5CO2 solution to turn it into a buffer with pH =4.63?
Chemistry
1 answer:
lubasha [3.4K]3 years ago
3 0

Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where  [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio  [A⁻]/HA].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2  can be determined.

So,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA]  = 4.63 - 4.20 =  log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 =  [A⁻]/[HA] = 2.69

 [A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

Molarity is moles per liter of solution, so we can calculate how many moles of  C6H5CO2⁻ the student needs to dissolve  in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L  = 0.34 mol C6H5CO2⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

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How much time would it take to distribute one avagadro's number of wheat grains if 10^10 grains are A)1.9*10^2 years B)1.9*10^10
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Answer:

It takes  1.9 \times 10^{6} years to distribute all the grains.

Explanation:

In one second, 10^{10} grains of wheat are distributed.

We are supposed to find the time it would take to Distribute Avogadro number of grains.

1 avogadro number = 6.022 \times 10^{23}

Number of grains distributed in 1 day = 86400 \times 10^{10}

                                                             = 8.64 \times 10^{14}

Number of grains distributed in 1 year= 8.64 \times 10^{14} \times 365

                                 = 3.1536 \times 10^{17}

Time taken = \frac{Number of grains to be distributed}{grains distributed in 1 year}

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Why must the mass of magnesium be less than 0.09 g?
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The energies of electrons are quantized. What does that mean?.
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The correct statement should be The energies of electron are <em>quantized</em> when <em>they are bounded to an atom.</em>

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<em><u>Thanks for joining brainly community! </u></em>

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A rubber ball is held 4.0 m above a concave spherical mirror with a radius of curvature of 1.5 m. At time equals zero, the ball
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The time elapsed when the ball placed above the concave mirror and the image formed would be at the same location is 0.55 s.

<h3>Image distance</h3>

The position of the image formed is determined using the followimg mirror formula;

\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

where;

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f = R/2

f = 1.5/2

f = 0.75 m

When the ball and its image is in the same position, u = v

The position of the ball is calculated as;

\frac{1}{f} = \frac{1}{u} + \frac{1}{u} \\\\\frac{1}{f} = \frac{2}{u} \\\\u = 2f\\\\u = 2(0.75)\\\\u = 1.5 \ m

<h3>Time of motion of the ball</h3>

The time taken for the ball to travel the caluclated distance is determined as;

h = ut + ¹/₂gt²

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1.5 = 4.9t²

t² = 1.5/4.9

t² = 0.306

t = 0.55 s

Thus, the time elapsed when the ball and its image are at the same location is 0.55 s.

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