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Karo-lina-s [1.5K]
3 years ago
13

When the hydronium ion concentration of a solution is increased by a factor of 10, it means that

Chemistry
1 answer:
Dahasolnce [82]3 years ago
4 0

Answer: The pH of the solution will decrease by 1 unit.

Explanation:

Let say the hydronium ion concentration of a solution is 0.001

pH = - log [H+]

= -log 1.0 x 10^3 = 3

if the hydronium ion concentration is increased by 10 , that is: 0.001 x 10 = 0.01

pH = -log [1.0 x 10^2]

= 2

Therefore, the pH is reduced by 1 unit when the hydronium ion concentration is increased by a factor of 10.

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For the following problem convert both the reactants to moles and balance chemical equationsThe reaction of 167 g Fe2O3 with 85.
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Let's start by balancing the reaction:

Fe_2O_3+CO\longrightarrow Fe+CO_2

As we can see, C appears only on two comopunds, CO and CO₂, and since both have 1 C each, their coefficients have to be the same for C to be balanced. However, CO has 1 O and CO₂ has 2, so there is a difference of 1 O betwee them.

The other source of O is Fe₂O₃, that has 3 O. So, we must choose a coefficient for CO and CO₂ such that the difference between the numbers of O is a multiple of 3, that way we can fix this difference with the O from Fe₂O₃. So, we can put coefficients of 3 on both of them:

Fe_2O_3+3CO\longrightarrow Fe+3CO_2

That way, we maintained C balanced (3 on each side) and now we have 3 + 3 O on the left side and 6 O on the right side, so the same amount.

Now, we just have to calance Fe, but it is easy since we have it alone in Fe. Since we have 2 on the left side, it is enough to put a coefficient of 2 on Fe to get the balanced reaction:

Fe_2O_3+3CO\longrightarrow2Fe+3CO_2

Now, to convert from mass to number of moles, we need the molar masses of the reactants, which we can calculate from the atomic weights of the elemnts in each of them:

M_{Fe_2O_3}=2\cdot M_{Fe}+3\cdot M_O=(2\cdot55.845+3\cdot15.9994)g/mol=159.6882g/molM_{CO}=1\cdot M_C+1\cdot M_O=(1\cdot12.0107+1\cdot15.9994)g/mol=28.0101g/mol

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\begin{gathered} M_{Fe_{2}O_{3}}=\frac{m_{Fe_2O_3}}{n_{Fe_{2}O_{3}}} \\ n_{Fe_2O_3}=\frac{m_{Fe_2O_3}}{M_{Fe_{2}O_{3}}}=\frac{167g}{159.6882g/mol}=1.045787\ldots mol \end{gathered}\begin{gathered} M_{CO}=\frac{m_{CO}}{n_{CO}} \\ n_{CO}=\frac{m_{CO}}{M_{CO}}=\frac{85.8g}{28.0101g/mol}=3.063180\ldots mol \end{gathered}

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\begin{gathered} Fe_2O_3\colon\frac{n_{Fe_2O_3}}{1}=\frac{1.045787\ldots mol}{1}=1.045787\ldots mol \\ CO\colon\frac{n_{CO}}{3}=\frac{3.063180\ldots mol}{3}=1.021060\ldots mol \end{gathered}

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