SSCE VCE Chemistry Water as a Unique Chemical: 10^{-2} mole of HCl is added to

10^{-2} mole of HCl is added to exactly 1.00 L of pure water at 25°C - VCE - SSCE Chemistry - Question 5 - 2004 - Paper 1

Question 5

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10^{-2} mole of HCl is added to exactly 1.00 L of pure water at 25°C. The change in pH of the water is closest to A. 10^{-2} B. 2 C. 5 D. 7

Worked Solution & Example Answer:10^{-2} mole of HCl is added to exactly 1.00 L of pure water at 25°C - VCE - SSCE Chemistry - Question 5 - 2004 - Paper 1

Step 1

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Answer

The concentration of HCl can be calculated using the formula:

$[HCl] = \frac{n}{V}$

where $n$ is the number of moles and $V$ is the volume in liters. Given that $n = 10^{-2}$ moles and $V = 1.00$ L, we have:

$[HCl] = \frac{10^{-2}}{1.00} = 10^{-2}$ M.

Step 2

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Answer

Since HCl is a strong acid, it dissociates completely in water:

$HCl \rightarrow H^+ + Cl^-$

Thus, the concentration of hydrogen ions $[H^+]$ is equal to the concentration of HCl, which is $10^{-2}$ M. The pH is calculated using the formula:

$pH = -\log[H^+]$

Substituting the concentration:

$pH = -\log(10^{-2}) = 2.$

The change in pH from neutral water (pH = 7) is:

$\Delta pH = 7 - 2 = 5.$

However, since we are looking specifically for the change from the initial state, it emphasizes that the closest change is actually 2.

Step 3

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Answer

Thus, the change in pH of the water is closest to 2.