Hi, I\'m stuck with solving problem #9 and #10. Would you show me how to solve t

Question

Hi,
I'm stuck with solving problem #9 and #10.

Would you show me how to solve this problem step-by-step please?

Thank you very much in advance!

nitrous acid (HNO2) formic acid (HCHO2) pka is considered as the best- puHer 1ooyate th 3.34 3.74 7.54 hypochlorous acid (HCLO) calculate the ratio of the conjugate base to the acid required to attain the desired P For the best choice, 8. Consider the titration of 25.0 ml of 0.100 M HCI with 0.100 M NaOH. Determine the pH after adding 30.0 mL of NaOH. Consider the titration of a 40.0 ml 0.10 M HNO2 (weak following: (a) The volume of KOlI solution required to reach the equivalence point. Number of md (b) The pH after adding 5.00 ml of KOH (c) The pH at one-half the equivalence point acid) solution with 0.20 M KOH. Determine the m perature No, of moles acid No of mole PC 10. Consider the following reaction d 2CH(g)C2H2(g) 3H2(8) te? A reaction mixture at 1700° initially contains (CH l0.115 M. At equilibrium, the mixture contains (CaHa = 0.035 M. What is the value of the equilibrium constant? yoles of HC 550m Remaining S-013: d mo 2.5 mol noles of No0 Total volum 25.0 30.0-55Dm lx 30 ml 03 mol nte Hal dissociate completelu 123

Explanation / Answer

9)

a) HNO2 + KOH ---> KNO2 + H2O

1 mol HNO2 = 1 mol KOH

No of mol of HNO2 taken = 40*0.1 = 4 mmol

No of mol of KOH required = 4 mmol

volume of of KOH required = 4/0.2 = 20 ml

b) pH of acidic buffer = pka + log(KNO2/KOH)

pka of HNO2 = 3.34

no of mol of HNO2 taken = 40*0.1 = 4 mmol

No of mol of KOH added = KNO2 formed = 5*0.2 = 1 mmol

pH = 3.34 + log(1/(4-1))

    = 2.86

c) at one half equivalence point, pH = pka , because at this point [KNO2] = [HNo2]

   pH = 3.34

10)

         2CH4(g) <=====> C2H2(g) + 3H2(g)

initial   0.115 M          -        -

change     -2x             +x       +3x

equilb    0.115-2x         x         3x

x = 0.035 M

so that, at equilibrium

[C2H2] = X = 0.035 M

[H2] = 3x = 3*0.035 = 0.105 M

[CH4] = 0.115-2*0.035 = 0.045 M

Kc = [C2H2][H2]^3/[CH4]^2

    = 0.035*0.105^3/0.045^2

    = 0.02

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