The pH of a solution prepared by mixing 45 mL of 0.183 M KOH and 65 mL of 0.145 M HCl is ?.
To find the pH of the solution, we need to calculate the concentration of H+ ions in the solution first.
Step 1: Write down the balanced chemical equation for the reaction between KOH and HCl.
KOH + HCl -> KCl + H2O
Step 2: Determine the number of moles for each reactant.
For KOH:
moles of KOH = concentration of KOH in M * volume of KOH in L
= 0.183 M * 0.045 L
= 0.008235 moles
For HCl:
moles of HCl = concentration of HCl in M * volume of HCl in L
= 0.145 M * 0.065 L
= 0.009425 moles
Step 3: Determine the limiting reactant.
The limiting reactant is the reactant that is present in the smallest amount. In this case, HCl has a smaller number of moles (0.009425 moles) compared to KOH (0.008235 moles). Therefore, HCl is the limiting reactant.
Step 4: Determine the moles of excess reactant.
To find the amount of KOH remaining after the reaction, we need to subtract the moles of HCl consumed from the initial moles of KOH.
moles of KOH remaining = initial moles of KOH - moles of HCl consumed
= 0.008235 moles - 0.009425 moles
= -0.00119 moles
Since we cannot have a negative amount of KOH remaining, it means that all the KOH reacted and there is no excess.
Step 5: Calculate the total volume of the solution.
total volume = volume of KOH + volume of HCl
= 45 mL + 65 mL
= 110 mL
= 0.11 L
Step 6: Calculate the concentration of H+ ions.
The moles of H+ ions produced in the reaction are equal to the moles of HCl consumed.
moles of H+ = 0.009425 moles
The concentration of H+ ions is given by the moles of H+ ions divided by the total volume of the solution.
concentration of H+ ions = moles of H+ / total volume
= 0.009425 moles / 0.11 L
= 0.08568 M
Step 7: Calculate the pH of the solution.
pH = -log[H+]
= -log(0.08568)
= 1.066
Therefore, the pH of the solution prepared by mixing 45 mL of 0.183 M KOH and 65 mL of 0.145 M HCl is approximately 1.066.
To determine the pH of the solution, we need to calculate the concentration of the resulting solution after mixing the KOH and HCl solutions. We can use the concept of stoichiometry and the concept of the concentration of an acid and a base to calculate the concentration of H+ in the solution. Once we have the concentration of H+, we can then calculate the pH using the formula pH = -log[H+].
First, let's calculate the moles of KOH and HCl for each solution:
For the KOH solution:
Moles of KOH = concentration (in M) x volume (in L)
= 0.183 M x (45 mL / 1000 mL/L)
= 0.008235 moles
For the HCl solution:
Moles of HCl = concentration (in M) x volume (in L)
= 0.145 M x (65 mL / 1000 mL/L)
= 0.009425 moles
Next, let's determine which component, KOH or HCl, is in excess. The component in excess is the limiting reactant that determines the outcome of the reaction:
To do this, we need to compare the amount of moles of KOH with the amount of moles of HCl. By examining the balanced chemical equation for the reaction between KOH and HCl:
KOH + HCl → KCl + H2O
We can see that the balanced equation has a 1:1 ratio between KOH and HCl, meaning that for every one mole of KOH, one mole of HCl is required.
Since the moles of KOH (0.008235 moles) are less than the moles of HCl (0.009425 moles), KOH is the limiting reactant.
Now, let's calculate the moles of HCl left after the reaction with KOH:
Moles of HCl left = Moles of HCl - Moles of KOH used
= 0.009425 moles - 0.008235 moles
= 0.00119 moles
To determine the concentration of H+ in the final solution, we need to find the total volume of the solution after mixing the KOH and HCl solutions:
Total volume of the solution = volume of KOH solution + volume of HCl solution
= 45 mL + 65 mL
= 110 mL
Next, let's calculate the concentration of H+ in the final solution:
Concentration of H+ = Moles of HCl left / Total volume of the solution (in L)
= 0.00119 moles / (110 mL / 1000 mL/L)
= 0.01082 M
Finally, let's calculate the pH using the formula pH = -log[H+]:
pH = -log(0.01082)
= 1.97
Therefore, the pH of the solution prepared by mixing 45 mL of 0.183 M KOH and 65 mL of 0.145 M HCl is approximately 1.97.
moles KOH = M x L = ??
moles HCl = M x L = ??
Which is in excess. by how much? That will be H^+ (if HCl is in excess) or OH^- (if KOH is in excess) and you determine pH from that final value.