what is the PH of a solution prepared by adding enough water to 15.00g of NaNO2 to make 275ml of solution
Calculate the molarity.
M = mols/L and mols = grams/molar mass
I get 15.00/69 = 0.217 mols
and M = 0.217/0.275 L = 0.790 M. You need to confirm this. I just estimated the molar mass.
The pH is determined by the hydrolysis (reaction with water) of the nitrite ion.
NO2^- + HOH ==> HNO2 + OH^-
Kb = Kw/Ka = (HNO2)(OH^-)/(NO2^-) = ??
You know Kw and ka for HNO2 so you can calculate Kb.
Now wet up an ICE table.
I = Initial concentration:
C = change in concentration:
E = equilibrium concentration.
Initial:
(NO2^-) = 0.790 M
(HNO2) = 0
(OH^-) = 0
change:
(OH^-) = +y
(HNO2) = +y
(NO2^-) = -y
Equilibrium:
(NO2^-) = 0.790 - y
(HNO2) = 0 + y = y
(OH^-) = 0 + y = y
Now substitute the equilibrium values above into the Kb expression above, and solve for y. That will be equal to the (OH^-), change that to pOH, then subtract from 14 to obtain pH.
Post your work if you get stuck.
thank you so much!
To calculate the pH of a solution prepared by dissolving NaNO2, we first need to find the concentration of the NaNO2 solution. We can use the formula:
Concentration (in moles per liter) = mass of solute (in grams) / molar mass of solute
Step 1: Calculate the moles of NaNO2:
Molar mass of NaNO2 = (22.99 g/mol) + (14.01 g/mol) + (2 * 16.00 g/mol) = 69.00 g/mol
Moles of NaNO2 = 15.00 g / 69.00 g/mol = 0.217 moles
Step 2: Calculate the concentration:
Volume of solution = 275 mL = 275/1000 L = 0.275 L
Concentration of NaNO2 = 0.217 moles / 0.275 L = 0.789 M
Step 3: Calculate the pOH:
pOH = -log10[OH-]
For NaNO2, the compound dissociates into Na+ and NO2-. Since NaNO2 is a salt, it does not significantly react with water to produce OH-. Therefore, the concentration of OH- is negligible, and we assume it to be zero.
pOH = -log10[OH-] = -log10(0) = Infinity
Step 4: Calculate the pH:
pH + pOH = 14
pH = 14 - pOH = 14 - Infinity = undefined
Therefore, the pH of the solution prepared by adding enough water to 15.00g of NaNO2 to make 275 mL of solution is undefined.
To determine the pH of a solution, we need to know the concentration of the relevant ions. In this case, we're dealing with sodium nitrite (NaNO2).
First, we need to calculate the molar mass of NaNO2:
NaNO2 = (22.99 g/mol * 1) + (14.01 g/mol * 1) + (16.00 g/mol * 2)
= 22.99 g/mol + 14.01 g/mol + 32.00 g/mol
= 69.00 g/mol
Next, we need to calculate the number of moles of NaNO2 present in the solution. We can use the formula: Moles = Mass / Molar mass.
Moles of NaNO2 = 15.00 g / 69.00 g/mol
≈ 0.2174 mol (rounded to 4 decimal places)
Now, we can calculate the molarity (M) of the NaNO2 solution. Molarity is defined as moles of solute divided by the volume of the solution in liters.
Molarity (M) = Moles / Volume (in liters)
= 0.2174 mol / 0.275 L
≈ 0.7909 M (rounded to 4 decimal places)
Since we know the concentration of the NaNO2 solution, we can calculate the pOH using the equation: pOH = -log10(OH^-). Assuming complete ionization, the concentration of OH^- ions is the same as the concentration of NaNO2.
pOH = -log10(0.7909 M)
≈ 0.103 (rounded to 3 decimal places)
Finally, we can subtract the pOH from 14 (pH + pOH = 14) to find the pH of the solution.
pH = 14 - 0.103
≈ 13.897 (rounded to 3 decimal places)
Therefore, the pH of the solution prepared by adding enough water to 15.00g of NaNO2 to make 275ml of solution is approximately 13.897.