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Dahasolnce [82]
2 years ago
6

Saccharin, a sugar substitute, is a weak acid with pka=2.32 at 25 ∘c. it ionizes in aqueous solution as follows: hnc7h4so3(aq)←−

→h+(aq)+nc7h4so−3(aq) part a what is the ph of a 0.11 m solution of this substance?
Chemistry
1 answer:
MrMuchimi2 years ago
3 0
Saccharin is considered as weak acid:
pH of weak acid = \frac{1}{2} pKa +  \frac{1}{2} pCa
pKa = 2.32 (given) and
pCa = -log (acid concentration) = - log (0.11) = 0.96
so pH = (\frac{1}{2}* 2.32 ) + ( \frac{1}{2} * 0.96) = 1.64
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There are:

3.41 moles of C

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3.40 moles of O.

Why?

To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

C_{6}H_{8}O_{6}

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.

We have that:

C_{6}=12.0107g*6=72.08g\\\\H_{8}=1.008g*8=8.064g\\\\O_{6}=15.999g*6=95.994g\\\\C_{6}H_{8}O_{6}=72.08g+8.064g+95.994g=176.138g

To know the percent of each element, we need to to the following:

C=\frac{72.08g}{176.138g}*100=0.409*100=40.92(percent)\\\\H=\frac{8.064g}{176.138g}*100=4.58(percent)\\\\O=\frac{95.994}{176.138g}*100=54.49(percent)

So, we know that for the 100 grams of the compound, there are:

40.92 grams of C

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We know the molecular masses of each element:

C=12.0107\frac{g}{mol}\\\\H=1.008\frac{g}{mol}\\\\O=15.999\frac{g}{mol}{mol}

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

C=\frac{40.92g}{12.010\frac{g}{mol}}=3.41mol\\\\H=\frac{4.58g}{1.008\frac{g}{mol}}=4.54mol\\\\O=\frac{54.49g}{15.999\frac{g}{mol}}=3.40mol

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.

Have a nice day!

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