Calculate the pH when 0.446 g of Cu(NO3)2 (s) are dissolved in 250.0 mL of water
Do you have any Ka or Kb values?
To calculate the pH when copper(II) nitrate (Cu(NO3)2) is dissolved in water, we need to follow these steps:
1. Write the balanced equation of the dissociation of copper(II) nitrate in water:
Cu(NO3)2 -> Cu2+ + 2NO3-
2. Calculate the concentration of Cu2+ ions in moles per liter (M):
To find the concentration, we need to know the number of moles of Cu(NO3)2 dissolved in water. First, we calculate the number of moles (n) using the formula:
n = mass / molar mass
The molar mass of Cu(NO3)2 can be calculated as follows:
Cu: 1 atom * atomic mass of Cu
N: 2 atoms * atomic mass of N
O: 6 atoms * atomic mass of O
Therefore, the molar mass of Cu(NO3)2 = (1 * atomic mass of Cu) + (2 * atomic mass of N) + (6 * atomic mass of O)
After calculating the molar mass, we can calculate the number of moles:
n = 0.446 g / molar mass
Now we calculate the concentration (C) in moles per liter (M):
C = n / volume
The volume (V) is given as 250.0 mL, so we need to convert it to liters:
V = 250.0 mL / 1000 (to convert mL to L)
3. Calculate the pOH:
The pOH is calculated using the formula:
pOH = -log10 (OH- concentration)
Since Cu(NO3)2 dissociates into Cu2+ and NO3- ions, and there are two NO3- ions for every one Cu2+ ion, the concentration of hydroxide ions (OH-) is zero in this case. Therefore, pOH = 0.
4. Calculate the pH:
The pH can be calculated using the equation:
pH + pOH = 14
Therefore, pH = 14 - pOH
Substituting the value of pOH as 0, we get:
pH = 14 - 0
Therefore, the pH is 14.
Note: In this case, the pH will be 14 as there are no hydroxide ions produced from the dissociation of Cu(NO3)2.