3. Dorji was provided with four colourless solutions. He was asked to

Chemistry

1 answer:

givi [52]3 years ago

5 0

Answer:

First, we can test Solution 1. We know that Sodium Hydroxide is a strong base. If we test acids on blue litmus paper, they will turn red. If we test bases on red litmus paper, they will turn blue. So, you can test all the of the solutions- water, sodium hydroxide and hydrochloric acid with blue and red litmus paper. HCl, Hydrochloric acid is an acid, so it will turn blue litmus paper red. It will not turn red litmus blue. The acids will turn blue litmus paper red. The bases will turn red litmus paper blue. Only water is a neutral liquid, which will not turn blue litmus paper red or red litmus paper blue. It will not change the colour of it. Thus, if you test all the solutions with blue and red litmus paper, you will know which solution is water. Water is the only one which is neutral. It is the only solution which cannot change the colour of any litmus paper. Thus, you can identify it very easily.

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The main acid in the stomach is hydrochloric acid (HCl). Which chemical equations show a reaction that occurs when someone takes

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Read 2 more answers

Write both answers to at least two decimal places. Calculate the pH of a 0.160 M solution of KOH.Part 2 (1 point) Calculate the

Evgen [1.6K]

To calculate the pH of a solution, we first need to find the concentration of hydronium ions in the solution. Since KOH is a strong base, it dissociates completely in water to produce hydroxide ions (OH-) and potassium ions (K+).

The concentration of hydronium ions in a solution of KOH can be calculated using the concentration of hydroxide ions and the equilibrium constant for water, which is equal to 1.00 x 10^-14 at 25 degrees Celsius.

The concentration of hydroxide ions in a 0.160 M solution of KOH is equal to the concentration of KOH, which is 0.160 M. The concentration of hydronium ions in the solution can be calculated using the equation below:

[H3O+] = (1.00 x 10^-14) / [OH-]

Substituting the concentration of hydroxide ions into the equation above, we get:

[H3O+] = (1.00 x 10^-14) / (0.160 M) = 6.25 x 10^-13 M

To calculate the pH of the solution, we need to take the negative logarithm of the concentration of hydronium ions. This can be done using the equation below:

pH = -log([H3O+])

Substituting the concentration of hydronium ions into the equation above, we get:

pH = -log(6.25 x 10^-13) = 12.20

The pH of a 0.160 M solution of KOH is 12.20.

To calculate the pOH of a solution, we first need to find the concentration of hydroxide ions in the solution. Since we already calculated this value above, we can simply use the concentration of hydroxide ions we found earlier: 0.160 M.

To calculate the pOH of the solution, we need to take the negative logarithm of the concentration of hydroxide ions. This can be done using the equation below:

pOH = -log([OH-])

Substituting the concentration of hydroxide ions into the equation above, we get:

pOH = -log(0.160 M) = 1.80

The pOH of a 0.160 M solution of KOH is 1.80.

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