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andreyandreev [35.5K]

3 years ago

5

Estimate the solubility of m(oh)2 in a solution buffered at ph = 7.0, 10.0, and 14.0.

1 answer:

kherson [118]3 years ago

4 0

<span>When M(OH)2 dissolves we have M(OH)2 which produces M2+ and 2OHâ’ pH + pOH=14 At ph =7; we have 7+pOH=14 pOH=14â’7 = 7 Then [OHâ’]=10^(â’pOH) [OH-] = 10^(-7) = 1* 10^(-7) At ph = 10. We have, pOH = 4. And [OH-] = 10^(-4) = 1 * 10^(-4) Finally ph = 14. We have, pOH = 0 And then [OH-] = 10^(-0) -----anything raised to zero power is 1, but (-0)... So [OH-] = 1</span>

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Solution :

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339 26.23

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Now the example of the first order kinetics w.r.t volumetric analysis is :

$k=\frac{2.303}{t} \log \left(\frac{v_{\infty}-v_0}{v_{\infty}- v_t}\right)$

Here, $v_{\infty}= \text{ volume at }\infty = 39.81$

$v_{t}= \text{ volume at time 't' } = 27.80$

$v_0$ = volume at time 0 = 0

Since the interval is not constant, we take the time interval as

$=\frac{903+1503+1801}{3}$

$=\frac{4207}{3}$

= 1402.3333

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$k=\frac{2.303}{1402} \log \left(\frac{39.81-0}{39.81-27.80}\right)$

$=(0.001643) \log \left(\frac{39.81}{10.01}\right)$

= 0.001643 x 0.52045

= 0.00082

$= 8.55 \times 10^{-4} \ sec^{-1}$

Therefore, the first order rate constant is k $= 8.55 \times 10^{-4} \ sec^{-1}$ .

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