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  1. The  [tex]H+\\[/tex] concentration is [tex]1.41 * 10^{-5}\\[/tex]

    Explanation:

    The pH of any solution is given by the equation

    [tex]pH = -log [H+][/tex]

    Where [tex]H+\\[/tex] is the hydrogen ion concentration

    Substituting the given values in above equation, we get -

    [tex]4.85 = -log[H+]\\H += 1.41 * 10^{-5}\\[/tex]

  2. [H⁺] = 0.920 MpH = 0.036[OH⁻] = 1.086x10⁻¹⁴ M

    Explanation:

    To calculate [H⁺] first we calculate the total number of H⁺ ions.

    H⁺ from HCl ⇒ 49.3 mL * 1.19 gSolution/mL * 38 gHCl / 100 gSolution = 22.29 g HCl

    22.29 g HCl ÷ 36.45 g/mol = 0.612 mol HCl = 0.612 mol H⁺

    H⁺ from HNO₃ ⇒ 19.5 mL * 1.42 gSolution/mL * 70 gHCl / 100 gSolution = 19.38 g HNO₃

    19.38 g HNO₃ ÷ 63 g/mol = 0.308 mol HNO₃ = 0.308 mol H⁺

    Total H⁺ moles = 0.612 + 0.308 = 0.920 mol H⁺

    Final Volume = 1.00 L

    [H⁺] = 0.920 mol / 1.0 L = 0.920 M

    Now we calculate pH:

    pH = -log [H⁺]pH = -log(0.920) = 0.036

    To calculate [OH⁻], we calculate pOH:

    pOH = 14 - pHpOH = 14 - 0.036 = 13.964

    pOH = 13.964 = -log[OH⁻]

    [tex]10^{-13.964}[/tex] = [OH⁻] = 1.086x10⁻¹⁴ M

  3. the concentration of H+ = 3 ×  [tex]10^{-3}[/tex]   M and the pH = 2.6 of the solution

    Explanation:

    Based on reduction potentials, hydrogen is a better reducing agent than copper, therefore copper([tex]Cu^{2+}[/tex]) is the cathode and hydrogen ([tex]H_{2}[/tex]) is the anode.

    Cathode reaction (reduction): [tex]Cu^{2+}(aq) + 2e^{-}[/tex]  ⇒   Cu(s)

    Anode reaction (oxidation) :  [tex]H_{2}(g)[/tex]     ⇒    [tex]2H^{+}(aq) + 2e^{-}[/tex]

    net reaction:   [tex]H_{2}(g) +[/tex]    [tex]Cu^{2+}(aq)[/tex]   ⇒  [tex]2H^{+}(aq) + Cu(s)[/tex]

    [tex]E_{0}cell = E_{0}cathode - E_{0}anode[/tex]

    E cathode = 0.337 v

    [tex]E_{0}cell = + 0.337 - 0 = 0.337[/tex]

    Q(reaction quotient) = [tex]\frac{[H^{+}]^{2} }{[Cu^{2+}]P_{H2} }[/tex]

    for 2 electrons [tex]Cu^{2+} = 1.00M[/tex] but [tex]H^{+}[/tex] is unknown. we solve this using hernst equation.

    [tex]E = E^{0} -\frac{0.0257}{n}ln\frac{[H^{+}]^{2} }{[Cu^{2+}]P_{H2} }[/tex]

    [tex]0.490 = 0.337 -\frac{0.0257}{2}ln\frac{[H^{+}]^{2} }{[1][1]}[/tex]

    [tex]ln{[H^{+}]^{2} } = -11.9[/tex]

    [tex]2ln{[H^{+}] } = -11.9[/tex]

    [tex]ln{[H^{+}] } = -5.95[/tex]

    [tex][H^{+}] = 3* 10^{-3} M[/tex]

    pH = 2.6

  4. The pH of a solution is measure of the acidity of a certain solution based from the concentration of the hydrogen ions. It is associated with the hydrogen ion dissolved in the solution. It is expressed as pH = -log [H+]. We calculate the concentration of the hydrogen ions from this expression.

     pH = -log [H+]
    1.65 = -log [H+]
    antilog [- 1.65] = [H+]
    [H+] = 10^-1.65
    [H+] = 0.0224 M

  5. pH → 7.46

    Explanation:

    We begin with the autoionization of water. This equilibrium reaction is:

    2H₂O  ⇄   H₃O⁺  +  OH⁻            Kw = 1×10⁻¹⁴         at 25°C

    Kw = [H₃O⁺] . [OH⁻]

    We do not consider [H₂O] in the expression for the constant.

    [H₃O⁺] = [OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷ M

    Kw depends on the temperature

    0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻]  → [H₃O⁺] = [OH⁻]         at 0°C

    √0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M

    - log [H₃O⁺] = pH

    pH = - log 3.46×10⁻⁸ → 7.46

  6. The pH of a solution is a value used to measure the acidity of a solution and also a measure of the hydrogen ion present in the solution. It is logarithmically related to the hydrogen ion concentration. It is expressed as:

    pH = -log [H+]
    4.85 = -log[H+]
    [H+] = 1.41x10^-5 M

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