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3 years ago

Calculate the pH of a 0.187 M solution of a base that has a Kb = 5.86x10^-6

Chemistry

1 answer:

xxTIMURxx [149]3 years ago

5 0

Answer:

$pH=11$

Explanation:

Hello there!

In this case, since the ionization of the described base is:

$base+H_2O\rightarrow OH^-+baseH^+$

Thus, the equilibrium expression in terms of the reaction extent is:

$Kb=5.86x10^{-6}=\frac{x^2}{0.187-x}$

Thus, by solving for x we obtain:

$5.86x10^{-6}({0.187-x})=x^2\\\\1.10x10^{-6}-5.86x10^{-6}x-x^2=0$

So the values of x are:

$x_1=0.0010M\\x_2=-0.0010M$

So the feasible answer is 0.0010 M, thus we compute the pOH:

$pOH=-log(0.0010M)=3$

And therefore the pH:

$pH=14-pOH=14-3\\\\pH=11$

Best regards!

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Classify each element. Note that another term for main group is representative, another term for semimetal is metalloid, and the

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The question is incomplete, here is the complete question:

Classify each element. Note that another term for main group is representative, another term for semi-metal is metalloid, and the inner transition metals are also called the lanthanide and actinide series.

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Answer:

Hafnium and tantalum are transition elements.

Americium is a inner transition element.

Indium, Selenium and Radon are main group elements.

Arsenic is a metalloid.

Explanation:

Main group elements are the elements which belong to s block and p block. They are also known as representative elements.

S-block elements are defined as the elements whose last electron enters s-sub shell. The general electronic configuration of these elements is $ns^{1-2}$

P-block elements are defined as the elements whose last electron enters p-sub shell. The general electronic configuration of these elements is $np^{1-6}$

Metalloids are defined as the elements which show intermediate properties between metals and non-metals. There are 7 metalloids in the periodic table. They are: Boron, Silicon, germanium, Arsenic, Antimony, Tellurium and Polonium.

Transition elements are known as d-block elements. D block elements are defined as the elements whose last electron enters d sub shell. The general electronic configuration of these elements is $[(n-1)d^{1-10}ns^{0-2}]$

Inner transition elements are known as (f block) elements. (F block) elements are defined as the elements whose last electron enters (f subshell). The general electronic configuration of these elements is $[(n-2)f^{1-14}(n-1)d^{0-1}ns^{2}]$ . They are also known as lanthanide and actinide series.

For the given elements:

Option 1: Hf

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As, the last electron is entering the d subshell, it is a transition element.

Option 2: Am

Americium is the 95th element of the periodic table having electronic configuration of $[Rn]5f^{7}6d^07s^2$

As, the last electron is entering the (f subshell), it is a inner transition element.

Option 3: In

Indium is the 49th element of the periodic table having electronic configuration of $[Kr]5s^25p^1$

As, the last electron is entering the p subshell, it is a main group element.

Option 4: Ta

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As, the last electron is entering the d subshell, it is a transition element.

Option 5: As

Arsenic is the 33rd element of the periodic table having electronic configuration of $[Ar]4s^24p^3$

As, the last electron is entering the p subshell, it is a main group element. It shows an intermediate property of metal and non-metal. Thus, it is a metalloid.

Option 6: Se

Selenium is the 34th element of the periodic table having electronic configuration of $[Ar]4s^24p^4$

As, the last electron is entering the p subshell, it is a main group element.

Option 7: Rn

Radon is the 86th element of the periodic table having electronic configuration of $[Xe]4f^{14}5d^{10}6s^26p^6$

As, the last electron is entering the p subshell, it is a main group element.

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