A 20.0-mL sample of 0.30 M HClO was titrated with 0.30 M NaOH. The following data were collected during the titration. Determine Ka for HClO. Data Provided: mL NaOH added pH 5.00 6.98 10.00 7.46 15.00 7.93 20.00 10.31 (Hint: Find the volume of NaOH at equivalence point)

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PhantomWisdom PhantomWisdom · Answer 1 · 2019-12-19T11:29:15+01:00

Answer:

[tex]K_a=3.46 \times 10^{-8}.[/tex]

Explanation:

We know at equivalent point,

[tex]M_1V_1=M_2V_2\\M_{acid}V_{acid}=M_{base}V_{base}\\0.30 \times20=0.30 \times V_{base}\\V_{base}=20 ml.[/tex]

So, equivalent point is obtained at 20 ml.

Therefore, half equivalent point is obtained at 10 ml.

Also , at 10 ml NaOH , pH= 7.46 .

We know,

[tex]pH=pK_a[/tex]

Also, [tex]pK_a=-\ln K_a[/tex]

Taking anti log both sides .

We get [tex]K_a=3.46 \times 10^{-8}.[/tex]

Hence, it is the required question.

andromache andromache · Answer 2 · 2022-04-12T12:32:21+02:00

The volume of NaOH at equivalence point should be considered as the [tex]3.46*10^-8[/tex][tex]3.46*10^-8[/tex].

SInce

we know that

[tex]m_1v_1 = m_2v_2\\\\0.30 * 0.20 = 0.30* v_2[/tex]

so v2 = 20 ml

Now

half equivalent point is obtained at 10 ml.

Also , at 10 ml NaOH , pH= 7.46 .

So

here the volume should be [tex]3.46*10^-8[/tex]

Hence, The volume of NaOH at equivalence point should be considered as the [tex]3.46*10^-8[/tex]

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