A solution of 0.178 M KOH(28.3 ml) is mixed with 28.9ml of 0.133 M HCl. Assuming that the final solution is the sum of the initial volumes, caculate:
a)the molarity of the K+ cation
b)the molatiry of the Cl- anion
c)the pH of the final solution
d)the pOH of the final solution
moles KOH = M x L = ?
moles HCl = M x L = ?
All K^+ come from KOH. All Cl^- come from HCl.
a. (K^+) = moles K^/L soln
b. (Cl^-) = moles Cl^-/L soln.
c. KOH + HCl ==> KCl + H2O
(H^+) or (OH^-) is determined by which reagent is in excess. Subtract the moles and see which is in excess and convert to pH or pOH.
Post your work if you get stuck.
To calculate the molarity of the K+ cation (a), the molarity of the Cl- anion (b), the pH of the final solution (c), and the pOH of the final solution (d), we can follow these steps:
Step 1: Determine the moles of KOH and HCl.
First, we calculate the moles of KOH and HCl using the given molarity and volume.
Moles of KOH = Molarity of KOH × Volume of KOH solution
= 0.178 M × 28.3 ml (converted to liters)
Moles of HCl = Molarity of HCl × Volume of HCl solution
= 0.133 M × 28.9 ml (converted to liters)
Step 2: Determine the change in moles of K+ and Cl- ions.
Since KOH is a strong base and HCl is a strong acid, they react in a 1:1 ratio. Therefore, the number of moles of K+ ion produced will be equal to the moles of KOH, and the number of moles of Cl- ion produced will be equal to the moles of HCl.
Step 3: Determine the total volume of the final solution.
The total volume of the final solution is the sum of the initial volumes of the KOH and HCl solutions, which is 28.3 ml + 28.9 ml.
Step 4: Calculate the molarity of K+ cation (a) and Cl- anion (b).
The molarity can be calculated by dividing the moles of each ion by the total volume of the final solution.
Molarity of K+ cation (a) = Moles of K+ / Total volume of final solution
Molarity of Cl- anion (b) = Moles of Cl- / Total volume of final solution
Step 5: Determine the pH of the final solution (c) and pOH of the final solution (d).
The pH and pOH can be calculated using the formulas:
pH = -log10 [H+]
pOH = -log10 [OH-]
Step 6: Convert pOH to pH.
Since pH + pOH = 14, we can subtract the pOH from 14 to obtain the pH.
By following these steps and plugging in the calculated values, you can find the answers to the questions (a), (b), (c), and (d).