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If Solution X has a pH of 3 and Solution Y has a pH of 6, we can conclude that A - VCE - SSCE Chemistry - Question 4 - 2014 - Paper 1
Question 4
If Solution X has a pH of 3 and Solution Y has a pH of 6, we can conclude that A. [$H^+$] in Solution X is 1000 times that of [$H^+$] in Solution Y. B. [$H^+$] in So... show full transcript
Worked Solution & Example Answer:If Solution X has a pH of 3 and Solution Y has a pH of 6, we can conclude that A - VCE - SSCE Chemistry - Question 4 - 2014 - Paper 1
Step 1
A. [$H^+$] in Solution X is 1000 times that of [$H^+$] in Solution Y.
Answer
To determine the concentration of hydrogen ions [], we use the formula:
[$H^+] = 10^{-pH}
For Solution X (pH = 3): [$H^+]_X = 10^{-3} ext{ mol/L}
For Solution Y (pH = 6): [$H^+]_Y = 10^{-6} ext{ mol/L}
Calculating the ratio:
\frac{[$H^+]_X}{[$H^+]_Y} = \frac{10^{-3}}{10^{-6}} = 10^{3} = 1000
Thus, this statement is true.
Step 2
B. [$H^+$] in Solution X is half that of [$H^+] in Solution Y.
Answer
From the previous calculation, we know:
[$H^+]_X = 10^{-3} ext{ mol/L} [$H^+]_Y = 10^{-6} ext{ mol/L}
To check if [] in Solution X is half that of Solution Y:
[$H^+]_X \neq \frac{1}{2}[$H^+]_Y
Thus, this statement is false.
Step 3
C. [$OH^-$] in Solution Y is twice that of [$OH^-$] in Solution X.
Answer
To find the [] concentration, we use the relationship:
[$OH^-] = \frac{K_w}{[$H^+]}
Where (ion-product constant of water) at 25°C is approximately .
For Solution X:
[$OH^-]_X = \frac{10^{-14}}{10^{-3}} = 10^{-11} ext{ mol/L}
For Solution Y:
[$OH^-]_Y = \frac{10^{-14}}{10^{-6}} = 10^{-8} ext{ mol/L}
Calculating the ratio:
\frac{[$OH^-]_Y}{[$OH^-]_X} = \frac{10^{-8}}{10^{-11}} = 10^{3} = 1000
Thus, this statement is false.
Step 4