50 mL of 1.50* 10^-2 M HI(aq) is mixed with 75.00 mL of 1.10* 10^-2 M KOH(aq).
What is the pH of the Final solution
First you must recognize that HI and KOH are strong acid + strong base respectively. The product is KI (a neutral salt) + H2O; therefore, the pH will depend entirely upon the reagent (HI or KOH) that is in excess. How do you determine the one in excess?
Calculate moles HI initially. That is M x L = ?
Calculate moles KOH initially. That is M x L = ?
Now subtract one from the other to see which one is in excess (obviously, the other one goes to zero since all of it will be used). Then moles/volume = M. Finally, if KOH is in excess, the pOH is the -log(OH^-) remaining or if HI is in excess, pH = -log(H^+) remaining.
Post your work if you get stuck.
To find the pH of the final solution, we first need to determine if a reaction occurs between the two solutions. In this case, HI (hydroiodic acid) is a strong acid and KOH (potassium hydroxide) is a strong base. When a strong acid and a strong base react, they neutralize each other to form water and a salt.
The balanced chemical equation for the reaction between HI and KOH is:
HI (aq) + KOH (aq) → H2O (l) + KI (aq)
Since HI and KOH react in a 1:1 ratio, the limiting reagent will determine the amount of products formed. To determine the limiting reagent, we can compare the number of moles of each reactant.
The number of moles for HI can be calculated using the formula:
moles = concentration (in M) × volume (in L)
For HI:
moles of HI = 1.50 × 10^-2 M × 50 mL / 1000 mL/L
moles of HI = 0.75 × 10^-3 mol
For KOH:
moles of KOH = 1.10 × 10^-2 M × 75 mL / 1000 mL/L
moles of KOH = 0.825 × 10^-3 mol
Comparing the moles of HI and KOH, we can see that HI is the limiting reagent since it has fewer moles.
Since HI is a strong acid, it will completely dissociate in water to form H+ ions. In this case, since the concentration of HI is 0.75 × 10^-3 M, the concentration of H+ ions will also be 0.75 × 10^-3 M.
To find the pH of the final solution, we can use the definition of pH, which is the negative logarithm of the H+ ion concentration.
pH = -log[H+]
Substituting the concentration of H+ into the equation, we have:
pH = -log(0.75 × 10^-3)
pH ≈ 3.12
Therefore, the pH of the final solution is approximately 3.12.