All categories

2 years ago

0.40 L of HNO3 is titrated to equivalence using 0.18 L of 0.1 M NaOH. What is the concentration of the HNO3 ?

1 answer:

3 0

This problem is providing us with the volume of nitric acid that is titrated with 0.18 L of 0.1-M sodium hydroxide and asks for the concentration of the acid. At the end, the result turns out to be 0.045M, according to the following.

<h3>Acid-base titrations:</h3>

In chemistry, acid-base titrations allow us to quantify the volume or concentration of an acid or base via the following equation:

$M_AV_A=M_BV_B$

Where the subscript A stands for the acid and B for the base; which means one can calculate any of the variables there by knowing the other three. This equation is based on the balanced neutralization chemical equation, which takes place between the acid and the base.

Thus, we can write the reaction between NaOH and HNO3 as:

$HNO_3+NaOH\rightarrow NaNO_3+H_2O$

In such a way, we can solve for the concentration of the acid as shown below:

$M_A=\frac{M_BV_B}{V_A} \\\\M_A=\frac{0.1M*0.18L}{0.40L} \\\\M_A=0.045M$

Learn more about titration: brainly.com/question/25485091

You might be interested in

Which is NOT a reason why organisms communicate with one another?

Tpy6a [65]

Answer:

C. to find housing.

Explanation:

i hope this helps :)

3 0

2 years ago

Read 2 more answers

How many milliliters of 0.0896M LiOH are required to titrate 25.0 mL of 0.0759M HBr to the equivalence point?

slavikrds [6]

Answer:

$V_{LiOH}=21.8mL$

Explanation:

Hello,

In this case, during titration at the equivalence point, we find that the moles of the base equals the moles of the acid:

$n_{LiOH}=n_{HBr}$

That it terms of molarities and volumes we have:

$M_{LiOH}V_{LiOH}=M_{HBr}V_{HBr}$

Next, solving for the volume of lithium hydroxide we obtain:

$V_{LiOH}=\frac{M_{HBr}V_{HBr}}{M_{LiOH}} =\frac{0.0759M*25.0mL}{0.0896M} \\\\V_{LiOH}=21.8mL$

Best regards.

8 0

2 years ago

I love u.

Norma-Jean [14]

Answer:

umm ok lol thx for the f r e e points

8 0

3 years ago

Read 2 more answers

A rectangular solid is 2.3 cm wide, 12.2 mm long and 0.75 inch thick. What is the volume of the piece?

Step2247 [10]

V=abc

a = 2,3cm
b=12,2mm = 1,22cm
c = 0,75inch = 1,905cm

V = 2,3cm*1,22cm*1,905cm ≈ 5,35cm³

5 0

2 years ago

After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of

SIZIF [17.4K]

Answer:

b) The dehydrated sample absorbed moisture after heating

Explanation:

a) Strong initial heating caused some of the hydrate sample to splatter out.

This will result in a higher percent of water than the real one, because you assume in the calculation that the splattered sample was only water (which in not true).

b) The dehydrated sample absorbed moisture after heating.

Usually inorganic salts may absorbed moisture from the atmosphere so this will explain the 13% difference between calculated water percent the real content of water in the hydrate.

c) The amount of the hydrate sample used was too small.

It will create some errors but they do not create a difference of 13% difference as stated in the problem.

d) The crucible was not heated to constant mass before use.

Here the error is small.

e) Excess heating caused the dehydrated sample to decompose.

Usually the inorganic compounds are stable in the temperature range of this kind of experiments. If you have an organic compound which retain water molecules you may decompose the sample forming volatile compounds which will leave crucible so the error will be quite high.

6 0

3 years ago