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erik [133]
3 years ago
6

What are the two species in highest concentration in a 0.1 M solution of lactic acid (CH3CH(OH)C00H), (Ka

Chemistry
1 answer:
rusak2 [61]3 years ago
6 0

Answer:

1.4 × 10^-4.

Explanation:

C3H6O3 + H2O <======> C3H5O3^- + H3O^+ ------------------------------------------(1).

So, from the question above we are given the following parameters or data which is going to help in solving this particular Question/problem;

=>concentration of the solution of lactic acid (CH3CH(OH)C00H) = 0.1 M and pH = 2.44.

Therefore, the concentration of the hydrogen ion[H^+} can be determined from the pH formula given below;

pH = - log { H^+}.

2.44 = - log { H^+}.

Therefore, {H^+} = 0.0036 M.

From the equation (1) given above, we have that the ratio for the equilibrium reaction is 1 : 1 : 1 :1. Therefore, molarity of C3H5O3^- = 0.0036 M and the molarity of C3H6O3 =( 0.1 - 0.0036 M) = 0.0964 M at equilibrium.

Hence, ka = {C3H5O3^-} { H3O^+} /{C3H6O3} = ( 0.0036 M)^2 /(0.0964 M) = 1.4 × 10^-4.

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Answer:

When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K

Explanation:

Step 1: Data given

A sample of sulfur hexafluoride gas occupies a volume of 5.10 L

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The volume will be reduced to 2.50 L

Step 2 Calculate the new temperature via Charles' law

V1/T2 = V2/T2

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⇒with V2 = the reduced volume of the gas = 2.50 L

⇒with T2 = the new temperature = TO BE DETERMINED

5.10 L / 471 K = 2.50 L / T2

T2 = 2.50 L / (5.10 L / 471 K)

T2 = 230.9 K = -42.1

When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K

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