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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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A saturated solution of baso4 has a concentration of 0.5mol/l. a 55ml sample is taken by you. what is the mass of baso4 in the s
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Answer:

6.4 g BaSO₄

Explanation:

You have been given the molarity and the volume of the solution. To find the mass of the solution, you need to (1) find the moles BaSO₄ (via the molarity ratio) and then (2) convert moles BaSO₄ to grams BaSO₄ (via the molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given values.

Molarity (mol/L) = moles / volume (L)

(Step 1)

55 mL / 1,000 = 0.055 L

Molarity = moles / volume                             <----- Molarity ratio

0.5 (mol/L) = moles / 0.055 L                        <----- Insert values

0.0275 = moles                                             <----- Multiply both sides by 0.055

(Step 2)

Molar Mass (BaSO₄): 137.33 g/mol + 32.065 g/mol + 4(15.998 g/mol)

Molar Mass (BaSO₄): 233.387 g/mol

0.0275 moles BaSO₄          233.387 g
---------------------------------  x  -------------------  =  6.4 g BaSO₄
                                                1 mole

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