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ziro4ka [17]
3 years ago
7

A chemist is using a solution of HNO3 that has a pH of 3. 75. What is [H ] for the solution? Round to the nearest hundredth. × 1

0n M n =.
Chemistry
1 answer:
Alexeev081 [22]3 years ago
8 0

The estimation of the acidic and the basic character of a solution is called and defined by the pH. The \rm [H^{+}] concentration of the solution will be \rm 1.77\times 10^{-4}.

<h3>What is the pH?</h3>

pH is a measurement of the free hydrogen and hydroxyl ions in the solution of the substance.

It can be calculated as:

\rm pH  = - \rm log[H^{+}]

Given,

pH = 3.75

Substituting values in the equation:

\begin{aligned}\rm 3.75 &= -\rm log[H^{+}]\\\\\rm [H^{+}] &= 10^{-3.75}\\\\&= 1.7782 \times 10^{-4}\end{aligned}

Therefore,  \rm 1.77\times 10^{-4} is the concentration of the hydrogen ions.

Learn more about pH here:

brainly.com/question/10285267

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Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K
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Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

And the undergoing chemical reaction:

MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

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