Calculate the pH and the pOH for each of the following solutions:
1. A solution made by adding 4.67 grams of potassium hydroxide to enough water to make 5.00 L of final solution.
2.A solution made by adding 50 mL of 6.0 M hydrochloric acid to enough water to produce 2.0 L of final solution.
3. A solution made by adding 30.0 mL of 1.00 M nitric acid to 100 mL of water.Assume additive volum
1.mols KOH = grams/molar mass = 4.67/56.1 = 0.0832
M KOH = mols/L solution = 0.0832/5.00 = ? = (KOH) = (OH^-)
pOH = -log (KOH) = ?
Then pOH + pH = pKw = 1E-14. You know Kw and pOH, solve for pH.
2. 6 M HCl is being diluted so use the dilution formula.
mLa x Ma = mLb x Mb
50 mL x 6 M = 2000 mL x Mb
Solve for Mb which gives you the concentration of the "new" HCl. Then convert to pH from there since (H^+) = (HCl). Convert that to pOH.
3. Done the same way as #2.
NOTE: Since all of these (KOH, HCl, HNO3) are strong electrolytes, the (OH^-) or (H^+) is the same as (KOH) or (HCl) or (HNO3).
Post your work if you get stuck.
To calculate pH and pOH for each solution, we need to follow a few steps.
Step 1: Calculate the concentration of the solute in each solution.
For solution 1:
We are given the mass of potassium hydroxide (KOH) and the volume of the final solution. To find the concentration, we need to convert the mass to moles using the molar mass of KOH (56.11 g/mol):
moles of KOH = mass / molar mass
moles of KOH = 4.67 g / 56.11 g/mol
Once we have the moles of KOH, we can calculate the concentration:
concentration (in mol/L) = moles of KOH / volume of solution (in L)
For solution 2:
We are given the volume (in mL) and concentration (in M) of hydrochloric acid (HCl). We need to convert the volume to liters:
volume of HCl (in L) = 50 mL / 1000 mL/L
Now we can directly use the given concentration.
For solution 3:
We are given the volume (in mL) and concentration (in M) of nitric acid (HNO3). We need to convert both the volume and concentration to liters:
volume of HNO3 (in L) = 30 mL / 1000 mL/L
concentration (in M) = 1.00 M
Step 2: Calculate the pOH for each solution.
pOH is calculated using the formula:
pOH = -log10[OH-]
For solution 1:
Since we have potassium hydroxide (KOH), it dissociates in water to produce hydroxide ions (OH-). The balanced chemical equation for this dissociation is:
KOH -> K+ + OH-
Since the concentration of OH- is equal to the concentration of KOH, we can directly use the concentration of KOH to calculate pOH.
For solution 2:
Hydrochloric acid (HCl) is a strong acid that completely dissociates in water to produce hydrogen ions (H+). Since pOH is the negative logarithm of the hydroxide ion concentration, and HCl does not produce hydroxide ions, pOH will be 0.
For solution 3:
Nitric acid (HNO3) is also a strong acid that dissociates in water to produce hydrogen ions (H+). Similar to solution 2, there are no hydroxide ions produced, so pOH will be 0.
Step 3: Calculate the pH for each solution.
pOH and pH are related by the equation:
pH + pOH = 14
Using this equation, we can calculate pH for each solution.
Now, let's calculate the pH and pOH for each solution:
For solution 1:
Calculate the concentration of KOH:
concentration = 4.67 g / 56.11 g/mol / 5.00 L = ...
Use the concentration to calculate pOH:
pOH = -log10[OH-] = ...
Calculate pH:
pH = 14 - pOH = ...
For solution 2:
Since pOH is 0, pH will be 14.
For solution 3:
Since pOH is 0, pH will be 14.