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A titration the 200.0 mL of 1.00 M H2A to be done through 1.38 M NaOH. For the diprotic mountain H2A, Ka1 = 2.5 10–5, Ka2 = 3.1 10–9.
1.00 M H2A
1.38 m NaOH
Titration = 200.0 mL
Calculate moles of NaOH
calculate moles of H2A
therefore the mole of acid left = mole of H2A - mole of NaOH
= 0.667 - 0.46 = 0.207
pka = - log( ka )
= - log in ( 2.5 * 10^-5 ) = 4.61
calculate PH after ~ 100 ml of 1.38 M NaOH have actually been added
PH = pka + log
= 4.61 + log
Well the thedesigningfairy.com is 207.2. You take the lowest sig figs the you are using which would be 2500= 2 sig figs. 207.2 rounded to 2 sig figs would be 210
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The price of reaction in terms of the "rate regulation expression" includes the rate continuous (kk), the concentration that the reactants,
In this case, considering the given certain orders of reaction, we can write the rate regulation as:
Thus, considering the concentration of A and B to be 0.200 M and also 0.350 M respectively and also a price of 0.060 M/s, the rate consistent turns out:
This is what we understand as a third-order reaction due to the fact that the particular orders both include to 3.