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Define pH. Calculate the pH of: (i) a 0.01 M sodium hydroxide (NaOH) solution - Leaving Cert Chemistry - Question 7 - 2012

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Define pH. Calculate the pH of: (i) a 0.01 M sodium hydroxide (NaOH) solution. (ii) a solution of hydrochloric acid (HCl) of concentration 3.65 grams per litre. (b... show full transcript

Worked Solution & Example Answer:Define pH. Calculate the pH of: (i) a 0.01 M sodium hydroxide (NaOH) solution - Leaving Cert Chemistry - Question 7 - 2012

Step 1

Define pH.

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Answer

pH is defined as the negative logarithm of the concentration of hydrogen ions in a solution, mathematically represented as:

pH=extlog[H+]pH = - ext{log} [H^+]

Where ([H^+]) is the concentration of hydrogen ions. The pH scale typically ranges from 0 to 14, where a pH less than 7 indicates acidity, a pH of 7 indicates neutrality, and a pH greater than 7 indicates alkalinity.

Step 2

Calculate the pH of: (i) a 0.01 M sodium hydroxide (NaOH) solution.

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Answer

To calculate the pH of the sodium hydroxide solution, we first determine its pOH:

pOH=extlog[OH]pOH = - ext{log} [OH^-]

Given that NaOH dissociates completely in water, ([OH^-] = 0.01 M):

pOH=extlog(0.01)=2pOH = - ext{log} (0.01) = 2

Now, we find the pH using the relationship ( pH + pOH = 14 ):

pH=14pOH=142=12pH = 14 - pOH = 14 - 2 = 12

Step 3

Calculate the pH of: (ii) a solution of hydrochloric acid (HCl) of concentration 3.65 grams per litre.

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Answer

First, we need to convert grams per litre to molarity. The molar mass of HCl is approximately 36.46 g/mol.

Calculating the number of moles in 3.65 grams:

extMolesofHCl=3.65extg36.46extg/mol0.1extmol ext{Moles of HCl} = \frac{3.65 ext{ g}}{36.46 ext{ g/mol}} \approx 0.1 ext{ mol}

This gives us a concentration of approximately 0.1 M. As HCl is a strong acid, it dissociates completely in water:

Thus, ([H^+] = 0.1 M).

Now, we can calculate pH:

pH=extlog[H+]=extlog(0.1)=1pH = - ext{log} [H^+] = - ext{log} (0.1) = 1

Step 4

What substance would you add to the water (i) if the pH needed to be raised?

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To raise the pH of water, an alkaline substance such as lime (calcium hydroxide) can be added. The reaction results in the neutralization of excess acidity, thus increasing the pH.

Step 5

What substance would you add to the water (ii) if the pH needed to be lowered?

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To lower the pH of water, an acidic substance such as sulfuric acid can be added. This helps in neutralizing alkaline water, thus decreasing the pH.

Step 6

Name a flocculating agent and explain how it helps in the purification process.

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Answer

A common flocculating agent is aluminum sulfate (alum). It helps in the purification process by causing suspended particles in the water to clump together, forming larger aggregates that can easily be removed through sedimentation or filtration.

Step 7

What is meant by hardness of water?

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Hardness of water refers to the presence of dissolved minerals, primarily calcium and magnesium ions. It affects soap's lathering ability and contributes to scale formation in pipes and boilers.

Step 8

Distinguish between temporary and permanent hardness in water.

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Temporary hardness is caused by dissolved bicarbonate minerals that can be removed by boiling. In contrast, permanent hardness is due to sulfate or chloride compounds that cannot be removed by boiling and require chemical treatment to soften the water.

Step 9

Suggest one advantage and one disadvantage of hard water.

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Answer

One advantage of hard water is that it provides essential minerals like calcium and magnesium, which are beneficial for human health. However, one disadvantage is that it can lead to scale buildup in plumbing and appliances, reducing their efficiency and lifespan.

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