What volume of a 0.150 M solution of KOH must be added to 450.0 mL of the acidic solution to completely neutralize all of the acid? Express the volume in liters to three significant figures.

Full Question;

What volume of a 0.150 M solution of KOH must be added to 450.0 mL of the acidic solution of 300ml of 0.450M HCL to completely neutralize all of the acid? Express the volume in liters to three significant figures.

0.9l

Explanation:

First thing's first, we have to write out the balanced chemical equation.

KOH(aq) + HCl(aq) → KCl(aq) + H2O(l)

Potassium hydroxide, KOH, and hydrochloric acid, HCl, react in a 1:1 mole ratio to produce aqueous potassium chloride, KCl, and water.

From the reaction;

Na = Nb

Where Na = Number of moles of acid

Na = Ca * Va = 0.450 * 0.300 = 0.135

Nb = Cb * Vb = Cb * 0.150

Na = Cb * 0.150

0.135 = Cb * 0.150

Cb = 0.135 / 0.150 = 0.9L

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

po-215 i got it right on the quiz