Calculate the ph of the resulting solution if 34.0 ml of 0.340 m hcl(aq is added to

Question

Calculate the ph of the resulting solution if 34.0 ml of 0.340 m hcl(aq is added to

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2022-05-21T02:55:25+00:00 2 Answers 0

Answers ( 2 )

  1. Skylar
    0
    2022-05-21T02:57:00+00:00

    a. pH = 12.711

    b. pH = 1.769

    Further explanationn

    pH is the degree of acidity of a solution that depends on the concentration of Hu207a ions. The greater the value the more acidic the solution and the smaller the pH.n

    pH = – log [Hu207a]n

    So that the two quantities between pH and [Hu207a] are inversely proportional because they are associated with negative values.n

    A solution whose value is different by n has a difference in the concentration of Hu207a ion of 10u207f.n

    The pH value of a reaction between strong acid HCl and strong base NaOH can be estimated from the rest of the reaction productn

    1. If the remainder of the reaction product is obtained the remaining strong base of NaOH then the pH is sought from the concentration of OHu207b / [OHu207b] by using the formula pH = 14-pOHn

    [OH u207b] = b. Mb wheren

    b = number of OHu207bn

    Mb = strong base concentrationn

    2. If the rest of the reaction results obtained the remaining strong acid HCl, the pH is sought from the concentration of Hu207a / [Hu207a] by using the formulan

    [Hu207a] = a. Mn

    a = valence of acid / amount of Hu207a releasedn

    M = acid concentration.n

    3. if both of them run out of reaction, then pH = 7n

    n

    n

    We complete the answer choicesn

    Calculate the pH of the resulting solution if 34.0 mL of 0.340 M HCL (aq) is added ton

    a) 44.0mL of 0.340 M NaOH (aq).n

    b) 24.0mL of 0.440 M NaOH (aq)n

    n

    a. mole HCl = 34.0mL x 0.340 M = 11.56 mmoln

    mol NaOH = 44.0 ml x 0.340 M = 14.96 mmoln

    The remaining moles: 14.96 – 11.56 = 3.4 mmoln

    [OH-] = mmol / total volumen

    [OH-] = 3.4 / 44 + 34n

    [OH-] = 0.0513n

    pOH = – log 5.13.10-2n

    pOH = 1,289n

    pH = 14-1.289 = 12.711

    b. mole HCl = 34.0mL x 0.340 M = 11.56 mmoln

    mol NaOH = 24.0mL x 0.440 M = 10.56 mmoln

    The remaining moles: 11.56 -11.56 = 1 mmoln

    [H+] = mmol / total volumen

    [H+] = 1 / 24 + 34n

    [H+] =0.017

    pH = – log 0.017

    pH = 1.769

    Learn moren

    the pH of a solutionn

    https://brainly.com/question/4039716n

    Calculate the pHn

    https://brainly.com/question/9278932n

    the pH of a 2.0 M solution of HClO4n

    https://brainly.com/question/1599662n

    Keywords : pH, acid, base, HCl,NaOH

  2. Josephine
    0
    2022-05-21T02:57:13+00:00

    According to the reaction equation:

    HCl(aq) + NaOH(aq)u2192 NaCl(aq) + H2O(l)

    a) Part 1):

    first, we need to get moles HCl = molarity * volume

    = 0.34m * 0.034

    = 0.01156 mol

    then, we need moles of NaOH = molarity * volume

    = 0.34 * 0.039

    = 0.01326 mol

    when NaOH excess:

    u2234 NaOH remaining = 0.01326 – 0.01156

    = 0.0017 mol

    when the total volume = 0.039 + 0.034

    = 0.073 L

    u2234[OH] = moles / total volume

    = 0.0017mol / 0.073 L

    = 0.0233 M

    u2234 POH = -u33d2[OH-]

    = -u33d20.0233

    = 1.63

    u2234 PH = 14- 1.63

    = 12.37

    b) part 2:

    as we got moles HCl = molarity * volume

    = 0.34 * 0.034

    = 0.01156 mol

    then moles NaOH = molarity * volume

    = 0.39 * 0.044

    = 0.01716 mol

    NaOH remaining = 0.01716 – 0.01156

    = 0.0056 mol

    when the total volume = 0.034 + 0.044

    = 0.078 L
    u2234[OH-] = 0.0056 / 0.078

    = 0.0718 M

    u2234 POH = -u33d2[OH-]

    = -u33d20.0718

    = 1.14

    u2234PH = 14 – POH

    = 14 -POH

    = 14 – 1.14

    = 12.86

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