If 59.0 mL of 0.40 M HI is mixed with 100. mL of 0.10 M KOH what is the resulting pH?
HI + KOH ==> KI + H2O
59.0 mL x 0.40 M = 23.6 millimoles HI.
100 mL x 0.10M = 10 mmol KOH.
That's an excess of HI by 23.6-10 = 13.6 mmoles HI
M = mmoles/mL = 13.6mmol/159 mL = ?M and convert to pH.
1.07
To find the resulting pH, we need to calculate the concentration of the resulting solution and then determine its pH using the concentration of hydronium ions ([H3O+]).
First, let's calculate the moles of each compound. The moles of HI can be found using the formula:
moles of solute = concentration × volume (in liters)
moles of HI = 0.40 M × 0.0590 L = 0.0236 moles
Similarly, the moles of KOH can be calculated as:
moles of KOH = 0.10 M × 0.100 L = 0.010 moles
Now, we need to determine which compound is the limiting reagent. The limiting reagent is the one that is completely consumed in the reaction and determines the amount of product formed. In this case, HI is the limiting reagent because it has the fewer number of moles.
Since the balanced chemical equation for the reaction between HI and KOH is:
HI + KOH -> H2O + KI
We can see that the reaction consumes 1 mole of HI for every 1 mole of KOH.
With 0.0236 moles of HI, we know that 0.0236 moles of KOH will react completely. Therefore, the remaining KOH after the reaction will be:
moles of KOH remaining = initial moles - reacted moles
= 0.010 moles - 0.0236 moles
= -0.0136 moles
Since we can't have negative moles, it means that KOH is completely consumed in the reaction. This also simplifies the calculation, as we will only have H3O+ ions remaining in the solution.
Now, let's calculate the concentration of hydronium ions ([H3O+]) to find the pH. Since the volume of the resulting solution is the sum of the initial volumes (0.0590 L + 0.100 L = 0.159 L), we have:
[H3O+] = moles of H3O+ / volume of solution
= 0.0236 moles / 0.159 L
= 0.148 M
Finally, to find the pH, we can use the equation:
pH = -log[H3O+]
pH = -log(0.148)
= 0.83
Therefore, the resulting pH is approximately 0.83.