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The equivalence point for the titration of a 25.00 -mL sample of CsOH solution with 0.1062 $\mathrm{M} \mathrm{HNO}_{3}$ is at 35.27 $\mathrm{mL}$ . What is the concentration of the CsOH solution?

0.149 $\mathrm{M}$

Chemistry 101

Chapter 18

Representative Metals, Metalloids, and Nonmetals

Nonmetals Chemistry

University of Central Florida

Brown University

Lectures

02:44

A 25.00-mL sample of an un…

02:03

If 26.5 $\mathrm{mL}$ of a…

01:48

A 20.0 -mL sample of 0.115…

23:41

Find the pH of the equival…

01:08

A $10.00 \mathrm{mL}$ samp…

03:03

An HNO $_{3}$ solution has…

06:52

At the end point of a titr…

00:41

A $25.00 \mathrm{mL}$ samp…

01:21

A 25.00 -mL sample of an u…

01:31

Calculate the molarity of …

0:00

A 30.00-mL sample of an un…

06:56

Calculate the concentratio…

02:16

What is the molar concentr…

01:54

The concentration of a sod…

01:09

03:24

Consider the titration of …

02:32

In a titration experiment,…

01:38

Sketch a titration curve f…

01:51

01:24

What is the molarity of a …

question 66 is a strong acid, strong base titrate Asian calculation. So we need to know the story geometry of the Thai tray shin reaction in order to perform this calculation well, right. The caesium hydroxide reacting with nitric acid producing water and cesium nitrate, and we'll see that it is balanced as is and we no longer need. We don't need coefficients other than one to calculate the molar ity of caesium hydroxide solution. Polarity is equal to mold caesium hydroxide divided by leader solution. We already know the leader solution. It's a 25 mL solution of caesium hydroxide, or 250.25 leaders, so we have to calculate the moles caesium hydroxide. We can get the mold caesium hydroxide by starting with the volume of nitric acid 35.27 mil leaders or 0.35 to 7 leaders. We multiply that by the molar ity of nitric acid to get the moles nitric acid. We then look at the Stoke Eom, a tree of the balanced reaction and we see that one mole nitric acid reacts with one mole caesium hydroxide. Now we have moles, caesium hydroxide, all we need to do is divide this by the leaders caesium hydroxide, 25 mL, or 250.0 to 500 leaders, and we get a concentration of caesium hydroxide of 5000.1498 Moeller.

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