Question

calculate the poh and ph of a 0.002 m solution of potassium hydroxide (koh). potassium hydroxide is a strong base.
poh = -logoh⁻
poh = -log__________
poh = __________

ph + poh = 14.00
ph = ________ - ________
ph = __________

find the hydroxide concentration oh⁻ of a solution of hydrofluoric acid (hf) with a ph of 5.3.
poh = 14.00 - ph
poh = ________ - ________
poh = __________

oh⁻ = 10^(-poh)
oh⁻ = __________
oh⁻ = ________ * ________

Explanation:

Step1: Calculate pOH of KOH solution

KOH is a strong base, so \([OH^-] = 0.002\ M = 2\times10^{-3}\ M\).
\(pOH = -\log[OH^-] = -\log(2\times10^{-3})\)
\(= -\log(2) - \log(10^{-3}) \approx -0.3010 + 3 = 2.699\)

Step2: Calculate pH of KOH solution

Using \(pH + pOH = 14.00\),
\(pH = 14.00 - pOH = 14.00 - 2.699 = 11.301\)

Step3: Calculate pOH of HF solution

Given \(pH = 5.3\),
\(pOH = 14.00 - pH = 14.00 - 5.3 = 8.7\)

Step4: Calculate \([OH^-]\) of HF solution

Using \([OH^-] = 10^{-pOH}\),
\([OH^-] = 10^{-8.7} = 10^{-8 - 0.7} = 10^{-8} \times 10^{-0.7} \approx 10^{-8} \times 0.2 = 2\times10^{-9}\ M\)

Filling the blanks (KOH part):

• \(pOH = -\log[\boldsymbol{0.002}]\) (or \(2\times10^{-3}\))
• \(pOH = \boldsymbol{2.70}\) (rounded)
• \(pH = \boldsymbol{14.00} - \boldsymbol{2.70}\)
• \(pH = \boldsymbol{11.30}\)

Filling the blanks (HF part):

• \(pOH = \boldsymbol{14.00} - \boldsymbol{5.3}\)
• \(pOH = \boldsymbol{8.7}\)
• \([OH^-] = 10^{\boldsymbol{-8.7}}\)
• \([OH^-] = \boldsymbol{2\times10^{-9}}\ M\) (or \(1.995\times10^{-9}\ M\))

Answer:

For KOH solution:
\(pOH = -\log[0.002] \approx 2.70\), \(pH = 14.00 - 2.70 = 11.30\)

For HF solution:
\(pOH = 14.00 - 5.3 = 8.7\), \([OH^-] = 10^{-8.7} \approx 2\times10^{-9}\ M\)

(Blanks filled as per above calculations.)