Chemistry College

A 25.0 mL sample of H2SO4 requires 20.0 mL of 2.00 M KOH for complete neutralization. What is the molarity of the acid?

Answers

Answer 1

We must first figure out how many moles of acid the KOH neutralised before we can calculate the molarity of the acid. Using the balanced equation for the neutralisation reaction, we may accomplish this:

K2SO4 + 2H2O = H2SO4 2KOH

According to this equation, complete neutralisation requires two moles of KOH for every mole of H2SO4. We can use the following equation to get the quantity of H2SO4:

(volume H2SO4 x molarity H2SO4 / 1000) = moles H2SO4

We are aware of the H2SO4 volume (25.0 mL), the KOH volume (20.0 mL) required for neutralisation, and the molarity of the KOH (2.00 M). Rearranging the equation above will allow us to get the molarity of the H2SO4.we can rearrange the equation above and substitute in the known values:

molarity H2SO4 = (1000 x moles H2SO4) / volume H2SO4

molarity H2SO4 = (1000 x (20.0 mL x 2.00 M) / (2 x 25.0 mL))

molarity H2SO4 = 8.00 M

So, the molarity of the H2SO4 is 8.00 M.

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Related Questions

The following items are required to create a concise informative plot for each of the terms listed. (Some terms are used more than once.)
The choices are (reaction, chemicals or system being investigated / special conditions of the experiment / best fit line or curve with an equation / units (if any) / table with headers containing units / name or symbol of the variable)

Answers

Combustion of a hydrocarbon fuel in air is the reaction that is being studied. Chemicals: The two chemicals utilised in the experiment are the air and the hydrocarbon fuel. Special circumstances: Various pressures

Chemicals are substances with a particular composition and set of characteristics. They can exist as solids, liquids, gases, and even plasma, among other forms. Chemicals are essential to many sectors of the economy, including industry, healthcare, agriculture, and energy. They go into the creation of goods like plastics, fertilisers, medicines, and fuels. Chemicals can be utilised in industrial operations including material cleaning, purification, and treatment. To avoid harming both people and the environment, it's crucial to handle and utilise chemicals carefully. This covers employing personal protective equipment and handling and disposing of chemicals in the right ways. chemicals Chert, limestone, and banded iron

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A 25.00 gram sample of an unknown metal initially at 99.0 degrees Celsius is added to 50.00 grams of water initially at 14.33 degrees Celsius. The final temperature of the system is 20.15 degrees Celsius. Calculate the specific heat of the metal. (The specific heat of water is 4.184 J/g*C). Record your answer in scientific notation using three significant figures.

Answers

The specific heat of the metal can be calculated using calorimetric equation. The specific heat of the metal here is 0.61 J/g °C.

What is specific heat capacity?

The specific heat capacity of  substance is the heat energy required to raise its temperature by one degree Celsius per one gram  of the substance.

The calorimetric equation connecting the heat energy q, mass m, specific heat c and temperature difference ΔT is:

q = m c  ΔT

Here, the heat released from the metal is equal to the heat absorbed by water.

Therefore,

q metal = q water. Let c  be the specific heat of the metal.

25 g  × ( 99 - 20.15°C) ×c = 50 g  × ( 20.15°C- 14.33) × 4.18 J/g °C

                                         = 1217.5 J

Then c = 1217.5 J/25 g  × ( 99 - 20.15°C)  = 0.61 J/g °C.

Therefore, the specific heat of the metal is 0.61 J/g °C.

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The following questions pertain to the element located in period 4 and group 15 of the periodic table.

What is the name of this element?
What is the atomic symbol for this element?
Is this element a metal, a nonmetal, or a metalloid?
What is the atomic number of this element?
How many protons does a neutral atom of this element have?
How many electrons does a neutral atom of this element have?
What is the average number of neutrons in a neutral atom of this element?
How many valence electrons does a neutral atom of this element have?
Assuming the atom is at ground state, or its lowest energy level, how many energy levels exist in its electron cloud?
Is the ion of this element larger or smaller than its neutral atom?
What is the charge on the monatomic ion of this element?
How many protons are in a monatomic ion of this element?
How many total electrons are in a monatomic ion of this element?
How many valence electrons are in a monatomic ion of this element?

Answers

Answer:

The name of this element is Nitrogen.

The atomic symbol for this element is N.

Nitrogen is a nonmetal.

The atomic number of Nitrogen is 7.

A neutral atom of Nitrogen has 7 protons.

A neutral atom of Nitrogen has 7 electrons.

The average number of neutrons in a neutral atom of Nitrogen is 7.

A neutral atom of Nitrogen has 5 valence electrons.

Assuming the atom is at ground state, or its lowest energy level, it has 2 energy levels in its electron cloud.

A Nitrogen ion is the same size as a neutral Nitrogen atom.

The charge on the monatomic ion of Nitrogen is +3.

There are 7 protons in a monatomic ion of Nitrogen.

There are 10 total electrons in a monatomic ion of Nitrogen.

There are 7 valence electrons in a monatomic ion of Nitrogen.

A student wishes to determine the concentration of Ag+ (aq) in a solution of AgNO3(aq). The student combines 10.00 mL of AgNO3 (aq) with excess Na2SO4 (aq) and observes the formation of a white precipitate. The formation of the precipitate is represented by the following equation. 2 AgNO3 (aq) + Na2SO4 (aq) Ag2SO4 (s) + 2 NaNO3 (aq) (a) Write the balanced net ionic equation for the precipitation reaction. The student collects the precipitate by filtration and measures the mass of the filter paper and precipitate every 10 minutes as it dries. The student records the data in the following table. Mass of dry filter paper 0.88 g Mass of filter paper and precipitate immediately after filtration 4.82 g Mass of filter paper and precipitate after 10 minutes 4.37 g Mass of filter paper and precipitate after 20 minutes 4.01 g Mass of filter paper and precipitate after 30 minutes 3.79 g (b) Use the data above to calculate the number of moles of Ag2SO4(s) (molar mass 311.8 g/mol) that precipitated. (c) Calculate the concentration of Ag+ in the original 10.00 mL solution of AgNO3 (aq). (d) The concentration of Ag+ (aq) determined by the student is significantly higher than the actual concentration of Ag*(ag). Based on the student's data table, identify an error in the experimental procedure that led to this result.

Answers

(a) 2 Ag+ (aq) + SO42- (aq) Ag2SO4 is the balanced net ionic equation for the precipitation reaction (s)

(b) To get the mass of the precipitate, we must first determine the number of moles of Ag2SO4 that precipitated. To do this, we must subtract the mass of the dry filter paper (0.88 g) from the mass of the filter paper and precipitate right away upon filtration (4.82 g).

4.82 g of precipitation divided by 0.88 g yields 3.94 g.

The mass of the precipitate can then be converted to moles using the molar mass of Ag2SO4 (311.8 g/mol):

Ag2SO4 moles are equal to 3.94 g/311.8 g/mol, or 0.0126 moles.

(c) We must first figure out how many moles of Ag+ ions were in the original 10.00 mL solution of AgNO3 (aq) in order to compute the concentration of Ag+ there. According to the balanced net ionic equation from (a), 1 mole of AgNO3 (aq) is used for every 2 moles of Ag+ ions that precipitate. Consequently, 0.0126 moles / 2 = 0.00630 moles of Ag+ ions were present in the initial 10.00 mL solution of AgNO3 (aq).

The mass of the original solution may then be determined using the volume (10.00 mL) and density (1.01 g/mL) of the solution:

Volume x Density = 10.00 mL x 1.01 g/mL = 10.1 g Mass of Original Solution

Ag+ concentration is calculated as follows: Ag+ concentration = Ag+ moles / original solution volume (in liters) = 0.00630 moles / 0.01 L = 63.0 mol/L

(d) The mass of the filter paper was not subtracted from the total mass of the filter paper and precipitate, causing the student to overestimate the mass of the precipitate and the number of moles of Ag2SO4 that precipitated. This, in turn, led to an overestimate of the concentration of Ag+ in the original solution.

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