Why is acid base titration a useful technique in chemistry

1 answer:

7 0

Answer: Titration is particularly useful if we want to find out the amount or concentration of a known acid or base in a given sample.

Explanation: Suppose you have just been handed a beaker containing a fixed volume of HCL solution. If you simply just know the volume of the solution, you can titrate it against a reagent (base) to figure out the exact concentration of the solution (also known as molarity)

The solution is placed in a flask for titration and a minute amount of indicator is then added into the flask. The reagent (base) is placed in a burette and slowly added to the solution and indicator mixture. The amount of reagent used is recorded when the indicator causes a change in the color of the solution. Using Stoichiometry, we are then able to calculate the concentration and moles of the HCL solution used up in the reaction.

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Which of the following is true for balancing equations? A. The number of products should be equal to the number of reactants. B.

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D.) There must be an equal number of atoms of each element on both sides of the equation

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2 years ago

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You carefully weigh out 16.00 g of CaCO3 powder and add it to 64.80 g of HCl solution. You notice bubbles as a reaction takes pl

bulgar [2K]

Answer: Mass of $CO_2$ produced in this reaction was 6.56 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$CaCO_3(s)+2HCl(aq)\rightarrow H_2O(l)+CO_2(g)+CaCl_2(aq)$

Mass or reactants = Mass of $CaCO_3$ + mass of $HCl$ = 16.00 + 64.80 = 80.80 g

Mass of products = mass of aqueous solution + mass of $CO_2$ + = 74.24 + x g

Mass or reactants = Mass of products

80.80 g = 74.24 + x g

x = 6.56 g

Thus mass of $CO_2$ produced in this reaction was 6.56 grams

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3 years ago

How many grams of oxygen gas (02) are needed to completely react with 9.30 moles

Luba_88 [7]

Answer:

223 g O₂

Explanation:

To find the mass of oxygen gas needed, you need to (1) convert moles Al to moles O₂ (via the mole-to-mole ratio from reaction coefficients) and then (2) convert moles O₂ to grams O₂ (via the molar mass). When writing your ratios/conversions, the desired unit should be in the numerator in order to allow for the cancellation of the previous unit. The final answer should have 3 sig figs because the given value (9.30 moles) has 3 sig figs.

4 Al + 3 O₂ ----> 2 Al₂O₃
^ ^

Molar Mass (O₂): 32.0 g/mol

9.3 moles Al 3 moles O₂ 32.0 g
------------------- x --------------------- x -------------------- = 223 g O₂
4 moles Al 1 mole

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What effect does rapid exhalation of CO2 during exercise have on the concentration of H2CO3 in the blood?

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If there is a constant loss of $CO_{2}~(carbon~dioxide)$ the concentration of $H_{2} CO_{3}~(carbonic~acid)$ will be affected and decrease since $CO_{2}$ is a main component of $H_{2}CO_{3}$ .