Can you help me answer questions 19-20 zoom in read the question QUESTION 18 Wha

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Can you help me answer questions 19-20 zoom in read the question QUESTION 18 What is the pri of a bulter prepared fram 033 M HCN and 109 M CN49 10 10 Write your answer using 2 places to the right of the decimal QUESTION 19 The equlibihum pi of an aquecus solution of weak acd is 3.15. if the invtial concentration of the weuk acid is 0.800 M, what is the Ka of the ac Sax 10 12x 103 20 x 106 63x 10 QUESTION 20 A student titrates 25.00 m of 0 3421 M NaCH with 0 1768 MHCL What is the pH of thtit atsion solution after the adtition of 25.00 m of rant T-25

Explanation / Answer

18.

Ka = 4.9 x 10-10

So, pKa = - log Ka
             = - log (4.9 x 10-10)
             = 9.31

Using Henderson-Hesselbalach equation

pH = pKa + log { [salt] / [acid] }
     = pKa + log { [NaCN] / [HCN] }
     = 9.31 + log (0.109 / 0.833)
     = 9.31 + log (0.130852341)
     = 9.31 + (-0.88)
     = 8.43

19.

Let the weak acid be HA

         HA             =        H+        +        A-
IC:             0.800                      0                    0
C:                 - x                     + x                 + x
EC:        0.800 – x                   x                    x

So, Ka = [H+] [A-]/ [HA]

Now,

pH = 3.15
or, - log[H+] = 3.15
or, log[H+] = -3.15
[H+] = 10-3.15
[H+] = 0.00071

So,
[A-] = [H+] = 0.00071
[HA] = 0.800 – 0.00071 = 0.79929

So,
Ka = [H+] [A-]/ [HA]
      = (0.00071) x (0.00071) / (0.79929)
     = 6.3 x 10-7

20.

The reaction equation for HCl and NaOH is expressed as

HCl   +   NaOH   ----->   NaCl   +   H2O

In the above reaction; HCl and NaOH reacts in 1:1 ratio.

Now,

25.00 mL of 0.3421 M NaOH
So, moles of NaOH = 0.3421 M x 0.025 L = 0.0085525 moles

25.00 mL of 0.1768 M HCl
So, moles of HCl = 0.1768 M x 0.025 L = 0.00442 moles

So, 0.00442 moles of HCl will neutralize 0.00442 moles of NaOH.
Hence, unreacted moles of NaOH = 0.0085525 moles – 0.00442 moles
                                                      = 0.0041325 moles

Total volume = 25 mL + 25 mL = 50 mL = 0.050 L

So, [NaOH] = 0.0041325 moles / 0.050 L = 0.08265 M = [OH-]

pOH = - log [OH-]
        = - log (0.08265)
       = 1.08

pH = 14 – pOH
     = 14 – 1.08
     = 12.92

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