If 0.630 grams of HNO3 (molecular weight 63.0) are placed in 1 liter of distilled water at 25 degrees C, what will be the pH of the solution? (Assume that the volume of the solution is unchanged by the addition of the HNO3.)
A. 0.01
B. 0.1
C. 1
D. 2
E. 3
whats the freaking answer
HNO3 is a strong acid; it ionizes 100%.
moles HNO3 = grams/molar mass.
M HNO3 = moles/L soln.
pH = -log(HNO3).
Thank you!
HNO3 is a strong acid; it ionizes 100%.
moles HNO3 = grams/molar mass.
0.63/63=0.01
M HNO3 = moles/L soln.
0.01/1=0.01
pH = -log(M HNO3).
-log(0.01) = 2
To find the pH of the solution, we need to calculate the concentration of H+ ions in the solution.
First, we need to determine the number of moles of HNO3 that are in 0.630 grams. To do this, we divide the mass of HNO3 by its molar weight:
Number of moles = 0.630 g / 63.0 g/mol = 0.01 mol
Since we are assuming that the volume of the solution is unchanged after adding HNO3, the concentration of HNO3 in the solution would be the same as the number of moles: 0.01 mol/L.
HNO3 is a strong acid and dissociates completely in water, therefore the concentration of H+ ions in the solution is also 0.01 mol/L.
Now, to find the pH of the solution we can use the formula:
pH = -log[H+]
Plugging in the value of [H+] = 0.01 mol/L, we get:
pH = -log(0.01) = -(-2) = 2
Therefore, the pH of the solution is 2.
The correct answer is option D. 2.