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xxMikexx [17]
3 years ago
8

Acetylsalicylic acid (aspirin), HC9H7O4, is the most widely used pain reliever and fever reducer. Find the pH of 0.035 M aqueous

aspirin at body temperature (Ka at 37°C = 3.6 × 10−4.)
Chemistry
1 answer:
KatRina [158]3 years ago
7 0

<u> </u> The pH of 0.035 M aqueous aspirin is 2.48

<u>Explanation:</u>

We are given:

Concentration of aspirin = 0.035 M

The chemical equation for the dissociation of aspirin (acetylsalicylic acid) follows:

               HC_9H_7O_4\rightleftharpoons H^++C_9H_7O_4^-

<u>Initial:</u>         0.035

<u>At eqllm:</u>    0.035-x        x         x

The expression of K_a for above equation follows:

K_a=\frac{[C_9H_7O_4^-][H^+]}{[HC_9H_7O_4]}

We are given:

K_a=3.6\times 10^{-4}

Putting values in above expression, we get:

3.6\times 10^{-4}=\frac{x\times x}{(0.035-x)}\\\\x=-0.0037,0.0033

Neglecting the value of x = -0.0037 because concentration cannot be negative

So, concentration of H^+ = x = 0.0033 M

  • To calculate the pH of the solution, we use the equation:

pH=-\log[H^+]

We are given:

H^+ = 0.0033 M

Putting values in above equation, we get:

pH=-\log(0.0033)\\\\pH=2.48

Hence, the pH of 0.035 M aqueous aspirin is 2.48

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Americans spend up to $100 billion annually for bottled water (41 billion gallons). The only beverages with higher sales are car
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<u>Answer:</u> C) be hypertonic to Tank B.

<u>Explanation: </u>

<u> The ability of an extracellular solution to move water in or out of a cell by osmosis</u> is known as its tonicity. Additionally, the tonicity of a solution is related to its osmolarity, which is the <u>total concentration of all the solutes in the solution. </u>

Three terms (hypothonic, isotonic and hypertonic) are used <u>to compare the osmolarity of a solution with respect to the osmolarity of the liquid that is found after the membrane</u>.  When we use these terms, we only take into account solutes that can not cross the membrane, which in this case are minerals.

  1. If the liquid in tank A has a lower osmolarity (<u>lower concentration of solute</u>) than the liquid in tank B, the liquid in tank A would be hypotonic with respect to the latter.
  2. If the liquid in tank A has a greater osmolarity (<u>higher concentration of solute</u>) than the liquid in tank B, the liquid in tank A would be hypertonic with respect to the latter.
  3. If the liquid in tank A has the same osmolarity (<u>equal concentration of solute</u>) as the liquid in tank B, the liquid in tank A would be isotonic with respect to the latter.

In the case of the problem, option A is impossible because the minerals can not cross the membrane, since it is permeable to water only. There is no way that the concentration of minerals decreases in tank A, so <u>the solution in this tank can not be hypotonic with respect to the one in Tank B. </u>

Equally, both solutions can not be isotonic and neither we can say that the solution in tank A has more minerals that the one in tank B because the liquid present in tank B is purified water that should not have minerals. Therefore, <u>options B and D are also not correct.</u>

Finally, the correct option is C, since in the purification procedure the water is extracted from the solution in tank A to obtain a greater quantity of purified water in tank B. In this way, the solution in Tank A would be hypertonic to Tank B.

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In the laboratory you dilute 2.49 ml of a concentrated 6.00 m hydrochloric acid solution to a total volume of 50.0 ml. what is t
Marizza181 [45]
Equation: M1V1 = M2V2

Where M = concentration & V = volume

Step 1: Write down what is given and what you are trying to find

Given: M1 = 6.00M, V1 = 2.49mL, and V2 = 50.0mL
Find: M2

Step 2: Plug in the values into the equation

M1V1 = M2V2
(6.00M)(2.49mL) = (M2)(50.0mL)

Step 3: Isolate the variable (Divide both sides by 50.0mL so M2 is by itself)

(6.00M)(2.49mL) / (50.0mL) = M2

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