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2 years ago

10

You have an aqueous solution with a ph of 8.0. You add sodium chloride to a concentration of 1 gram per 100 milliliters. What ha

ppens to the ph?

2 answers:

Marat540 [252]2 years ago

6 0

Answer:

It stays the same.

Explanation:

As seen in the question above, you are manipulating a solution that has pH 8, that is, this is a basic solution. However, you decide to add sodium chloride at a concentration of 1 gram per 100 milliliters. Sodium chloride has ph7, which means that it is a neutral substance, that is, it is not a basic or acidic molecule.

Because of its neutral character and concentration, sodium chloride will not change the pH of the solution where it was added. As a result, the solution will remain at pH 8.

zysi [14]2 years ago

4 0

Answer:

what is the ph of sodium chloride

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• We obtained the above 10.00-mL solution by diluting a stock solution using a 1.00-mL aliquot and placing it into a 25.00-mL vo

garik1379 [7]

Answer:

a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) 0.0035 mole

c) 0.166 M

Explanation:

Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.

The equation of the reaction is expressed as:

$H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O$

1 mole 3 mole

The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

$H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O$

10 ml 17.50 ml

(x) M 0.200 M

Molarity = $\frac{0.2*17.5}{1000}$

= 0.0035 mole

c) What was the molar concentration of phosphoric acid in the original stock solution?

By stoichiometry, converting moles of NaOH to H₃PO₄; we have

= $0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}$

= 0.00166 mole of H₃PO₄

Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:

Molar Concentration = $\frac{mole \ \ of \ soulte }{ Volume \ of \ solution }$

Molar Concentration = $\frac{0.00166 \ mole \ of \ H_3PO_4 }{10}*1000$

Molar Concentration = 0.166 M

∴ the molar concentration of phosphoric acid in the original stock solution = 0.166 M

6 0

2 years ago

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$

Thus it takes $0.0241 \; \text{L}$ of sodium hydroxide to produce this buffer solution.

6 0

2 years ago

The ________________and ______________ of magma cause igneous rocks to form.

LekaFEV [45]

Answer:

The solidifying/cooling and crystallizing of magma causes igneous rocks to form.

6 0

2 years ago

Convert 90.23 kPa into atm

svlad2 [7]

Answer:

The answer is

<h2>0.89 atm </h2>

Explanation:

To convert from kPa to atm we use the conversion

101.325 kPa = 1 atm

If

101.325 kPa = 1 atm

Then

90.23 kPa will be

$\frac{90.23}{101.325} \\ =0.89050086...$

We have the final answer as

<h3>0.89 atm</h3>

Hope this helps you

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2 years ago

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zimovet [89]

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2 years ago