Question

What is the pH of 1.0 L of water to which (a) 1.5 mL of 3.0 M HCl or (b) 1.5 mL of 3.0 M NaOH has been added?

Answer 1

Answer:

(a) The pH is 2.3

(b) The pHis 11.7

Explanation:

First, it is necessary to calculate the concentration of (a) HCl and (b) NaOH in the final solution.

$\frac{1.5mL.3moles}{1000mL} = 4.5x10^{-3} moles$

$4.5x10^{-3} moles$ are in 1.0L of water, thus the final solution of (a) HCl and (b) NaOH have a concentration of $4.5x10^{-3} M$.

(a) HCl is a strong acid, so the concentration of protons in the solution is the same of HCl. To calculate the pH used the equation,

$pH = -Log [H^{+} ] = - Log (4.5x10^{-3}moles) = 2.3$

(b) NaOH is a strong base, so the concentration of oxydriles is the same of NaOH. To calculate the pOH used the equation,

$pOH = -Log [OH^{-} ] = - Log (4.5x10^{-3}moles) = 2.3$

Also, pH + pOH = 14 so,

pH = 14 - pOH = 14 - 2.3 = 11.7

Answer 2

Answer:

(a) 2.3

(b) 11.7

Explanation:

(a)

The moles of HCl in 1.5 mL of 3.0 M HCl are:

1.5 × 10⁻³ L × 3.0 mol/L = 4.5 × 10⁻³ mol

When 1.5 × 10⁻³ L of this solution are added to 1.0 L of water, the volume of the solution is:

1.0 L + 1.5 × 10⁻³ L = 1.0015 L

The concentration of HCl in this solution is:

4.5 × 10⁻³ mol/1.0015 L = 4.5 × 10⁻³ M

HCl is a strong monoprotic acid, so [H⁺] = 4.5 × 10⁻³ M. The pH is:

pH = -log [H⁺] = -log 4.5 × 10⁻³ = 2.3

(b)

The moles of NaOH in 1.5 mL of 3.0 M NaOH are:

1.5 × 10⁻³ L × 3.0 mol/L = 4.5 × 10⁻³ mol

When 1.5 × 10⁻³ L of this solution are added to 1.0 L of water, the volume of the solution is:

1.0 L + 1.5 × 10⁻³ L = 1.0015 L

The concentration of NaOH in this solution is:

4.5 × 10⁻³ mol/1.0015 L = 4.5 × 10⁻³ M

NaOH is a strong monohydroxide, so [OH⁻] = 4.5 × 10⁻³ M. The pOH is:

pOH = -log [OH⁻] = -log 4.5 × 10⁻³ = 2.3

The pH is

pH + pOH = 14.0

pH = 14.0 - pOH = 14.0 - 2.3 = 11.7