In a flask, 10.0 mL of a 0.100 M HCl solution is diluted to 1.00 L. In a second flask, 10.0 mL of a 0.100 M KOH solution is also diluted to 1.00 L. Which statement best describes the changes in pH in these flasks? pH change of the HCl solution pH change of the KOH solution A. increases by 2 B. increases by 2 C. decreases by 2 D. decreases by 2

SSCE VCE Chemistry Acid-Base Reactions: In a flask, 10.0 mL of a 0.100 M

In a flask, 10.0 mL of a 0.100 M HCl solution is diluted to 1.00 L - VCE - SSCE Chemistry - Question 7 - 2009 - Paper 1

Question 7

In a flask, 10.0 mL of a 0.100 M HCl solution is diluted to 1.00 L. In a second flask, 10.0 mL of a 0.100 M KOH solution is also diluted to 1.00 L. Which statement ... show full transcript

Worked Solution & Example Answer:In a flask, 10.0 mL of a 0.100 M HCl solution is diluted to 1.00 L - VCE - SSCE Chemistry - Question 7 - 2009 - Paper 1

Step 1

pH change of the HCl solution

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Answer

When 10.0 mL of 0.100 M HCl is diluted to 1.00 L, the concentration decreases significantly. The formula for pH is given by:

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

Initially, the number of moles of HCl is:

ext{moles of HCl} = 0.100 imes 0.010 = 0.001 ext{ moles}

When diluted to 1.00 L, the concentration becomes:

[ ext{H}^+] = rac{0.001 ext{ moles}}{1.00 ext{ L}} = 0.001 ext{ M}

Calculating the initial pH:

ext{pH}_{ ext{initial}} = - ext{log}(0.100) = 1.00

New pH after dilution:

ext{pH}_{ ext{final}} = - ext{log}(0.001) = 3.00

Thus, the pH increases by 2 units.

Step 2

pH change of the KOH solution

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Answer

Similarly, when 10.0 mL of 0.100 M KOH is diluted to 1.00 L, we analyze the change in pH. The initial moles of KOH are:

ext{moles of KOH} = 0.100 imes 0.010 = 0.001 ext{ moles}

After dilution to 1.00 L, the concentration of OH⁻ becomes:

[ ext{OH}^-] = rac{0.001 ext{ moles}}{1.00 ext{ L}} = 0.001 ext{ M}

Calculating the initial pOH:

ext{pOH}_{ ext{initial}} = - ext{log}(0.100) = 1.00

New pOH after dilution:

ext{pOH}_{ ext{final}} = - ext{log}(0.001) = 3.00

Using the relationship between pH and pOH, we find:

ext{pH} + ext{pOH} = 14 ightarrow ext{pH}_{ ext{final}} = 14 - 3 = 11

Thus, the pH decreases by 2 units.

In a flask, 10.0 mL of a 0.100 M HCl solution is diluted to 1.00 L - VCE - SSCE Chemistry - Question 7 - 2009 - Paper 1

Worked Solution & Example Answer:In a flask, 10.0 mL of a 0.100 M HCl solution is diluted to 1.00 L - VCE - SSCE Chemistry - Question 7 - 2009 - Paper 1

pH change of the HCl solution

pH change of the KOH solution

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