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.
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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
To determine the pH change of the HCl solution, we first calculate the initial concentration before dilution:
extInitialconcentrationofHCl=0.100extM
Diluating 10.0 mL to 1.00 L:
ext{Final concentration} = rac{10.0 ext{ mL} imes 0.100 ext{ M}}{1000 ext{ mL}} = 0.00100 ext{ M}
The pH of the original solution is:
extpH=−extlog(0.100)=1.00
The pH of the diluted solution is:
extpH=−extlog(0.00100)=3.00
Thus, the change in pH is:
3.00−1.00=2.00
This means that the pH of the HCl solution increases by 2.
Step 2
pH change of the KOH solution
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Answer
Next, we determine the pH change of the KOH solution:
extInitialconcentrationofKOH=0.100extM
Diluating 10.0 mL to 1.00 L:
ext{Final concentration} = rac{10.0 ext{ mL} imes 0.100 ext{ M}}{1000 ext{ mL}} = 0.00100 ext{ M}
The pOH of the original solution is:
extpOH=−extlog(0.100)=1.00
The pOH of the diluted solution is:
extpOH=−extlog(0.00100)=3.00
Then, we convert pOH to pH using the relation:
extpH+extpOH=14
Hence, the original pH of KOH is:
extpH=14−1.00=13.00
And the diluted pH is:
extpH=14−3.00=11.00
Thus, the change in pH is:
11.00−13.00=−2.00
This indicates that the pH of the KOH solution decreases by 2.
