A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 13.0 mL of KOH. Express your answer numerically.

pH= 1.17

Explanation:

The neutralization reaction between HBr (acid) and KOH (base) is given by the following equation:

HBr(aq) + KOH(aq) → KBr(aq) + H₂O(l)

According to this equation, 1 mol of HBr reacts with 1 mol of KOH. Then, the moles can be expressed as the product between the molarity of the acid/base solution (M) and the volume in liters (V). So, we calculate the moles of acid and base:

Acid:

M(HBr) = 0.15 M = 0.15 mol/L

V(HBr) = 50.0 mL x 1 L/1000 mL = 0.05 L

moles of HBr = M(HBr) x V(HBr) = 0.15 mol/L x 0.05 L = 7.5 x 10⁻³ moles HBr

Base:

M(KOH) = 0.25 M = 0.25 mol/L

V(HBr) = 13.0 mL x 1 L/1000 mL = 0.013 L

moles of HBr = M(HBr) x V(HBr) = 0.25 mol/L x 0.013 L = 3.25 x 10⁻³ moles KOH

Now, we have: 7.5 x 10⁻³ moles HBr > 3.25 x 10⁻³ moles KOH

HBr is a strong acid and KOH is a strong base, so they are completely dissociated in water: the acid produces H⁺ ions and the base produces OH⁻ ions. So, the difference between the moles of HBr and the moles of KOH is equal to the moles of remaining H⁺ ions after neutralization:

moles of H⁺ = 7.5 x 10⁻³ moles HBr - 3.25 x 10⁻³ moles KOH = 4.25 x 10⁻³ moles H⁺

From the definition of pH:

pH = -log [H⁺]

The concentration of H⁺ ions is calculated from the moles of H⁺ divided into the total volume:

total volume = V(HBr) + V(KOH) = 0.05 L + 0.013 L = 0.063 L

[H⁺] = (moles of H⁺)/(total volume) = 4.25 x 10⁻³ moles/0.063 L = 0.067 M

Finally, we calculate the pH after neutralization:

pH = -log [H⁺] = -log (0.067) = 1.17

11 players on a field that has a maximum of 11 players

the answer is 303.15 or 303