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3 years ago

LE

Chemistry

1 answer:

iVinArrow [24]3 years ago

3 0

Answer:

pH=2.16

Explanation:

Hello,

In this case, the first step is to compute the molarity of the 0.10%-w/v valproic acid solution by assuming 100 mL of solution:

$M=\frac{0.10g}{100mL}*\frac{1mol}{144.21g} *\frac{1000mL}{1L}\\ \\M=6.934x10^{-3}M$

Moreover, the equilibrium expression for the valproic acid (a weak one) is written as follows:

$Ka=\frac{[H^+][A^-]}{[HA]}$

Whereas HA represents the valproic acid and A⁻ its conjugate base. Thus, by computing Ka via its pKa and writing the aforementioned equilibrium expression in terms of $x$ we obtain:

$10^{-4.8}=1.585x10^{-5}=\frac{x*x}{6.934x10^{-3}-x}$

Thus, solving for $x$ , which also equals the concentration hydrogen ions, we obtain:

$x=[H^+]=6.93399x10^{-3}M$

Therefore, the pH is:

$pH=-log([H^+])=-log(6.93399x10^{-3})\\\\pH=2.16$

Best regards.

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Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 1.000 mg of caffeine

Yuliya22 [10]

Answer:

194 g/mol

Explanation:

1) Content of C:

All the C atoms in the 1.000 mg of caffeine will be found in the 1.813 mg of CO₂.

Mass of C in 1.813 mg of CO₂

Since the atomic mass of C is 12.01 g/mol and the molar of of CO₂ is 44.01, there are 12 mg of C in 44 mg of CO₂ and you can set the proporton:

12.01 mg C / 44.01 mg CO₂ = x / 1.813 g CO₂

⇒ x = 1.813 × 12.01 / 44.01 g of C = 0.49475 mg of C

Number of moles of C

number of moles = mass in g / atomic mass = 0.49475×10⁻³ g / 12.01 g/mol = 4.1195×10⁻⁵ moles = 0.041195 milimol

2) Content of H

All the H atoms in the 1.000 mg of caffeine will be found in the 0.4639 mg of H₂O

Mass of H in 0.4639 mg of H₂O

Since the atomic mass of H is 1.008 g/mol and the molar of of H₂O is 18.015 g/mol, there are 2×1.008 mg of H in 18.015 mg of H₂O and you can set the proporton:

2×1.008 mg H / 18.015 mg H₂O = x / 0.4639 mg H₂O

⇒ x = 0.4639 mg H₂O × 2 × 1.008 mg H / 18.015 mg H₂O = 0.051913 mg H

Number of moles of H

number of moles = mass in g / atomic mass = 0.051913 ×10⁻³ g / 1.008 g/mol = 5.1501× 10⁻⁵ moles = 0.051501 milimol

3) Content of N

All the N atoms in the 1.000 mg of caffeine will be found in the 0.2885 mg of N₂

Mass of N in 0.2885 mg of N₂ is 0.2885 mg

Number of moles of N

number of moles = mass in g / atomic mass = 0.2885 ×10⁻³ g / 14.007 g/mol = 2.0597× 10⁻⁵ moles = 0.020597 milimol

4) Content of O

The mass of O is calculated by difference:

Mass of O = mass of sample - mass of C - mass of H - mass of N

Mass of O = 1.000 mg - 0.49475 mg C - 0.051913 mg H - 0.2885 mg N

Mass of O = 0.1648 mg

Moles of O = 0.1648 × 10 ⁻³ g / 15.999 g/mol = 1.0303×10⁻⁵ mol = 0.01030 milimol

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Divide every number of mililmoles by the smallest number of milimoles:

C: 0.041195 / 0.01030 = 4
H: 0.051501 / 0.01030 = 5
N: 0.020597 / 0.01030 = 2

O: 0.01030 / 0.01030 = 1

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H: 5
N: 2
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4 × 12 + 5 × 1 + 2 × 14 + 1 × 16 = 97 g/mol

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