a large volume of 0.1590 m h2so3(aq) is treated with enough naoh(s) to adjust the ph of the solution to 5.63 . assuming that the addition of naoh(s) does not significantly affect the volume of the solution, calculate the final molar concentrations of h2so3(aq) , hso−3(aq) , and so2−3(aq) in solution given that the Ka1 and Ka2 values are 1.50×10−2 and 1.20×10−7 , respectively.

Answers

Answer 1

To calculate the final molar concentrations of H2SO3(aq), HSO−3(aq), and SO2−3(aq) in solution, we need to consider the dissociation of H2SO3. H2SO3(aq) can dissociate into HSO−3(aq) and H+(aq), and further into SO2−3(aq) and H+(aq).

Given that the Ka1 and Ka2 values are 1.50×10−2 and 1.20×10−7, respectively. Calculate the initial concentration of H2SO3(aq) using its volume and molarity. Use the Ka1 value to calculate the concentration of HSO−3(aq) and H+(aq) at equilibrium.

Subtract the concentration of H+(aq) from the initial concentration of H2SO3(aq) to find the final concentration of H2SO3(aq). Calculate the final concentration of HSO−3(aq) and SO2−3(aq) by subtracting the concentration of H+(aq) from their respective equilibrium concentrations.

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Related Questions

A graduated cylinder contains 26 cm3 of water. an object with a mass of 21 grams and a volume of 15 cm3 is lowered into the water. what will the new water level be

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When the object with a volume of 15 cm3 is lowered into the water in the graduated cylinder, the new water level will be 11 cm3.

The new water level in the graduated cylinder can be determined by considering the principle of displacement. When the object is lowered into the water, it will displace an amount of water equal to its own volume.

Given that the object has a volume of 15 cm3, it will displace 15 cm3 of water. Since the initial volume of water in the graduated cylinder is 26 cm3, the new water level can be calculated by subtracting the volume of water displaced by the object from the initial volume of water.

Therefore, the new water level in the graduated cylinder will be 26 cm3 - 15 cm3 = 11 cm3.

To summarize, when the object with a volume of 15 cm3 is lowered into the water in the graduated cylinder, the new water level will be 11 cm3.

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You prepare a stock solution that has a concentration of 2. 5 m. An aliquot with a volume of 10. 0 ml is removed from the solution. What is the concentration of the aliquot?.

Answers

The concentration of the aliquot is 2.5 M.

The concentration of a solution is defined as the amount of solute present per unit volume of the solution.

In this case, the stock solution has a concentration of 2.5 M (moles per liter).

An aliquot is a small portion or sample taken from a larger solution. In this scenario, an aliquot with a volume of 10.0 ml is removed from the stock solution.

Since the concentration of the stock solution is given in terms of moles per liter (M), the concentration of the aliquot will be the same as the concentration of the stock solution.

The concentration does not change when a specific volume is removed from the solution.

Therefore, the concentration of the aliquot is 2.5 M. It is important to note that the concentration remains the same regardless of the volume of the aliquot, as long as the proportion of solute to solvent remains constant.

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at constant temperature, a 144.0 ml sample of gas in a piston chamber has a pressure of 2.25 atm. calculate the pressure of the gas if this piston is pushed down hard so that the gas now has a volume of 36.0 ml.

Answers

The pressure of the gas would be 9.0 atm if the piston is pushed down hard to a volume of 36.0 ml.

To solve this problem, we can use Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume at constant temperature.

First, we need to set up the equation: P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

Given that the initial volume (V1) is 144.0 ml and the initial pressure (P1) is 2.25 atm, and the final volume (V2) is 36.0 ml, we can plug in the values into the equation:

2.25 atm * 144.0 ml = P2 * 36.0 ml

Next, we can solve for P2 by dividing both sides of the equation by 36.0 ml:

2.25 atm * 144.0 ml / 36.0 ml = P2

P2 = 9.0 atm

Therefore, the pressure of the gas would be 9.0 atm if the piston is pushed down hard to a volume of 36.0 ml.

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