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diamong [38]
2 years ago
6

Para-aminobenzoic acid (paba), p-h2nc6h4(cooh), is used in some sunscreens and hair conditioning products. Calculate the ph of a

n aqueous solution with [paba] = 0. 030 m and ka = 2. 2 × 10-5
Chemistry
1 answer:
never [62]2 years ago
4 0

the pH of an aqueous solution with [paba] = 0. 030 m and ka = 2. 2 × 10⁻⁵

<h3>What is pH ?</h3>

pH is a measure of how acidic/basic water is.

The range goes from 0 - 14, with 7 being neutral.

pHs of less than 7 indicate acidity, whereas a pH of greater than 7 indicates a base.

pH is really a measure of the relative amount of free hydrogen and hydroxyl ions in the water.

H₂NC₆H₄COOH    ---->     H₂NC₆H₄COO⁻ +  H⁺

 0.030M                                0                       0

 0.030-x                                x                       x

\rm K_{a}= \dfrac{[H^{+}][H_{2}NC_{6}H_{4}COO^{-} ]}{[H_{2}NC_{6}H_{4}COOH]}

= x² / 0.030-x

x = 8.01 * 10⁻⁴

[H⁺] = x = 8.01 * 10⁻⁴

pH = -log[ H⁺]

=-log[ 8.01 * 10⁻⁴]

pH = 3.09

Therefore the pH of an aqueous solution with [paba] = 0. 030 m and ka = 2. 2 × 10⁻⁵

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How many moles of LiOH are needed to react completely with 25.5 g of CO2
LekaFEV [45]

Answer:

3.18 mol

Explanation:

2LiOH+CO_{2}-> Li_{2}CO_{3} +H_{2}O

n(CO2) = mass/ Mr.

             = 25.5 / 16

             = 1.59 mol

As per the equation above,

n(LiOH) : n(CO2)

     2      :    1

∴  3.18   :  1.59

     

3 0
2 years ago
Hydrogen bonding occurs when hydrogen is bonded to N, O, or F. Which of the following molecules has hydrogen bonding, when bondi
Makovka662 [10]

Answer:

I think the answer is……

O B.H2S

Explanation:

I’m not sure tho, I’m just not 100% positive.

5 0
3 years ago
A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base
yawa3891 [41]

Answer:

11.39

Explanation:

Given that:

pK_{b}=4.82

K_{b}=10^{-4.82}=1.5136\times 10^{-5}

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{1.805\ g}{82.0343\ g/mol}

Moles= 0.022\ moles

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.022}{0.055}

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

                                  B +   H₂O    ⇄     BH⁺ +        OH⁻

At t=0                        0.4                          -              -

At t =equilibrium     (0.4-x)                        x           x            

The expression for dissociation constant is:

K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}

1.5136\times 10^{-5}=\frac {x^2}{0.4-x}

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³  M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>

5 0
2 years ago
Please help with these two (last question is nwse)
Vlad1618 [11]

Answer:

your answer gonna be 20 miles

5 0
2 years ago
Draw the major product for the reaction of 1-butyne with water in the presence of catalytic TfOH (i.e., CF3SO3H). Then answer th
Mademuasel [1]

Answer:

2-Butanone

Explanation:

From the given information:

The presence of mercury as an acid catalyst brings about the addition of water to the triple bond which yields enol. Then, according to Markownikov's rule and after tautomerism has occurred, we have a methyl ketone ( 2- Butanone) as the product.

The answer regarding the transformation is addition and hydration.

4 0
3 years ago
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