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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.

**thedesigningfairy.com:**

4.95

**Explanation:**

1.00 M H2A

1.38 m NaOH

Titration = 200.0 mL

Calculate moles of NaOH

=

= 0.46calculate moles of H2A

=

= 0.667therefore 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

=**4.95**

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,

**thedesigningfairy.com:**

**Explanation:**

Hello.

In this case, considering the given certain orders of reaction, we can write the rate regulation as:

**^2" alt="r=k**

**^2" align="absmiddle" class="latex-formula">**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:

**^2}\\ \\k=\frac0.060M/s(0.200M)(0.350M)^2 \\\\k=2.45M^-2s^-1" alt="k=\fracr**

**^2\\ \\k=\frac0.060M/s(0.200M)(0.350M)^2 \\\\k=2.45M^-2s^-1" align="absmiddle" class="latex-formula">**This is what we understand as a third-order reaction due to the fact that the particular orders both include to 3.