Calculate the pOH and the pH of a 5.0 x 10-2 M solution of NaOH.

pOH = 1.3, pH = 12.7

Explanation:

Since NaOH is a strong base, it will completely ionize; further, since it completely ionizes, our hydroxide concentration (a product of the ionization) will be the same as the given concentration of NaOH.

NaOH -> Na⁺ + OH⁻, [OH⁻] = 5.0 x 10^-2 M

pOH is the negative log of the hydroxide concentration, so plug our hydroxide concentration in:

pOH = -log[OH⁻] = -log[5.0 x 10^-2 M] = 1.3

Since pH + pOH = 14, we can plug in pOH and solve for pH:

pH + 1.3 = 14

pH = 14 - 1.3 = 12.7

Thus, our pOH = 1.3 and pH = 12.7.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol