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.
[tex]\frac{1.5mL.3moles}{1000mL} = 4.5x10^{-3} moles[/tex]
[tex]4.5x10^{-3} moles[/tex] are in 1.0L of water, thus the final solution of (a) HCl and (b) NaOH have a concentration of [tex]4.5x10^{-3} M[/tex].
(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,
[tex]pH = -Log [H^{+} ] = - Log (4.5x10^{-3}moles) = 2.3[/tex]
(b) NaOH is a strong base, so the concentration of oxydriles is the same of NaOH. To calculate the pOH used the equation,
[tex]pOH = -Log [OH^{-} ] = - Log (4.5x10^{-3}moles) = 2.3[/tex]
Also, pH + pOH = 14 so,
pH = 14 - pOH = 14 - 2.3 = 11.7
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
