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Acids, bases and pHChemistryCollegeabout 1:43

Find the pH of a 0.05 M weak base with Kb = 1.8e-05: animated lesson

A narrated, step-by-step acids, bases and ph video lesson with captions. Final answer: pH = 10.97, pOH = 3.027, [H⁺] = 1.064×10^-11 M, [OH⁻] = 9.397×10^-4 M (basic).

Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.6 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Worked example
    Find the pH of a 0.05 M weak base with Kb = 1.8e-05.
  2. 2. Steps, part 1
    1. Kb=x2C−xK_b = \frac{x^2}{C - x}
      Set up the equilibrium (ICE table)
    2. x2+Kx−KC=0⇒x=−K+K2+4KC2=9.397×10−4 Mx^2 + K x - K C = 0 \Rightarrow x = \frac{-K + \sqrt{K^2 + 4KC}}{2} = 9.397 \times 10^{-4}\ \text{M}
      Solve the quadratic exactly
    3. xC×100=1.879%\frac{x}{C} \times 100 = 1.879\%
      Percent ionisation
    4. pOH=−log⁡10(9.397×10−4)=3.027\text{pOH} = -\log_{10}(9.397 \times 10^{-4}) = 3.027
      pOH from [OH⁻]
  3. 3. Steps, part 2
    1. pH=−log⁡10[H+]=−log⁡10(1.064×10−11)=10.973\text{pH} = -\log_{10}[\mathrm{H^+}] = -\log_{10}(1.064 \times 10^{-11}) = 10.973
      pH
    2. pOH=14−pH=3.027,[OH−]=Kw[H+]=9.397×10−4 M\text{pOH} = 14 - \text{pH} = 3.027,\quad [\mathrm{OH^-}] = \frac{K_w}{[\mathrm{H^+}]} = 9.397 \times 10^{-4}\ \text{M}
      pOH and [OH⁻] (25 °C, Kw = 1.0 × 10⁻¹⁴)
  4. 4. The answer
    pH=10.97\text{pH} = 10.97

    pH = 10.973

  5. 5. Check
    • [H⁺][OH⁻] = 1×10^-14 ≈ 1.0×10^-14.
    • The solution is basic.

Transcript

The problem. Here is the problem: Find the pH of a 0.05 M weak base with Kb equals 1.8e minus 05. Convert between hydrogen ion concentration, hydroxide concentration, pH and pOH in water at 25 °C.

Steps, part 1. Set up the equilibrium (ICE table). x is the equilibrium concentration of OH to the power minus, which equals the amount that ionised. Solve the quadratic exactly. Percent ionisation. Below 5 percent, so the shortcut x approximately the square root of KC would also have been acceptable. pOH from [OH to the power minus ].

Steps, part 2. pH. pOH and [OH to the power minus ] (25 °C, Kw equals 1.0 times 10 to the power minus 14).

The answer. So the answer is pH equals 10.97, pOH equals 3.027, [H to the power plus ] equals 1.064 times 10 to the power negative 11 M, [OH to the power minus ] equals 9.397 times 10 to the power negative 4 M (basic). pH is a logarithmic scale: each unit is a tenfold change in hydrogen ion concentration. Below 7 the solution is acidic, above 7 it is basic, at 25 °C.

Check. [H to the power plus ][OH to the power minus ] equals 1 times 10 to the power negative 14 approximately 1.0 times 10 to the power negative 14. The solution is basic.

About acids, bases and ph

Find pH, pOH and ion concentrations of acidic and basic solutions.

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Find the pH of a 0.05 M weak base with Kb = 1.8e-05 (Lesson)