35.0 mL of a 0.250 M solution of KOH is titrated with 0.150 M HCl. After 35.0 mL of the HCl has been added, the resultant solution is:

Well, after adding 35.0 mL of the HCl, I hope you had your umbrella handy because things might have gotten a little acidic! The resultant solution is bound to be a bit sour, and definitely not recommended for a refreshing beverage.

To determine the resultant solution after the addition of 35.0 mL of 0.150 M HCl to 35.0 mL of 0.250 M KOH, we need to consider the reaction between these two compounds.

The balanced chemical equation for the reaction between KOH and HCl is:

KOH + HCl -> KCl + H2O

From the balanced equation, we can see that KOH reacts with HCl in a 1:1 ratio. This means that one mole of KOH reacts with one mole of HCl to produce one mole of KCl and one mole of water.

First, let's calculate the number of moles of HCl added to the solution:

Number of moles of HCl = Volume of HCl (in L) * Concentration of HCl

Given that the volume of HCl added is 35.0 mL (which is equivalent to 0.0350 L) and the concentration of HCl is 0.150 M, we can calculate:

Number of moles of HCl = 0.0350 L * 0.150 mol/L = 0.00525 mol

Since the reaction is in a 1:1 molar ratio between KOH and HCl, the number of moles of KOH that reacted is also 0.00525 mol.

Next, let's calculate the remaining moles of KOH in the solution:

Number of moles of KOH remaining = Initial moles of KOH - Moles of KOH reacted

The initial moles of KOH can be calculated by multiplying the volume of KOH (35.0 mL or 0.0350 L) by the concentration of KOH (0.250 M):

Initial moles of KOH = 0.0350 L * 0.250 mol/L = 0.00875 mol

Number of moles of KOH remaining = 0.00875 mol - 0.00525 mol = 0.00350 mol

Finally, let's find the concentration of KOH in the resultant solution:

Concentration of KOH = Number of moles of KOH remaining / Volume of resultant solution

The volume of the resultant solution is the sum of the volumes of KOH and HCl added, which is 35.0 mL + 35.0 mL = 70.0 mL (0.0700 L).

Concentration of KOH = 0.00350 mol / 0.0700 L = 0.0500 M

Therefore, the resultant solution after the addition of 35.0 mL of 0.150 M HCl to 35.0 mL of 0.250 M KOH is a 0.0500 M solution of KOH.

Jake, follow the same procedure I showed you above on your NaOH/HCl titration. Determine if the acid or the base is in excess. That will determine whether the solution at that point is acidic or basic. Post your work if you get stuck.