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Anna007 [38]
3 years ago
14

What happens when acid and base are mixed together

Chemistry
1 answer:
saul85 [17]3 years ago
6 0
A chemical reaction happens if you mix together an acid and a base. The reaction is called neutralisation<span>, and a neutral solution is made if you add just the right amount of acid and base together.</span>
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Question 8 of 8&gt; 0 Attempt 2 A chemist needs to determine the concentration of a sulfuric acid solution by titration with a s
Nesterboy [21]

Answer:

12.29 M

Explanation:

  • The reaction that takes place is:

H₂SO₄ + 2NaOH → 2Na⁺ + SO₄⁻² + 2H₂O

  • Now let's calculate the <u>moles of H₂SO₄ that were titrated</u>:

0.1284 M * 0.1915L*\frac{1molAcid}{2molNaOH} = 0.01229 mol H₂SO₄.

  • Thus, the <u>concentration of the diluted solution is</u>:

0.01229 mol H₂SO₄ / 0.010 L = 1.229 M

  • Finally, the <u>concentration of the original acid solution is:</u>

1.229 M*\frac{250mL}{25mL} = 12.29 M

5 0
3 years ago
If you have access to stock solutions of 1.00 M H3PO4, 1.00 M of HCl, and 1.00 M NaOH solution, (and distilled water of course),
garri49 [273]

Answer:

0.10L of 1.00M of H₃PO₄ and 0.1613L of 1.00M NaOH

Explanation:

The pKa's of phosphoric acid are:

H₃PO₄/H₂PO₄⁻ = 2.1

H₂PO₄⁻/HPO₄²⁻ = 7.2

HPO₄²⁻/PO₄³⁻ = 12.0

To make a buffer with pH 9.40 we need to convert all H₃PO₄ to H₂PO₄⁻ and an amount of H₂PO₄⁻ to HPO₄²⁻

To have a 50mM solution of phosphoures we need:

2L * (0.050mol / L) = 0.10 moles of H₃PO₄

0.10 mol * (1L / mol) = 0.10L of 1.00M of H3PO4

To convert the H₃PO₄ to H₂PO₄⁻ and to HPO₄²⁻ must be added NaOH, thus:

H₃PO₄ + NaOH → H₂PO₄⁻ + H₂O + Na⁺

H₂PO₄⁻ + NaOH → HPO₄²⁻ + H₂O + Na⁺

Using H-H equation we can find the amount of NaOH added:

pH = pKa + log [A⁻] / [HA] <em>(1)</em>

<em>Where [A-] is conjugate base, HPO₄²⁻ and [HA] is weak acid, H₂PO₄⁻</em>

<em>pH = 7.40</em>

<em>pKa = 7.20</em>

[A-] + [HA] = 0.10moles <em>(2)</em>

Replacing (2) in (1):

7.40 = 7.20 + log 0.10mol - [HA] / [HA]

0.2 = log 0.10mol - [HA] / [HA]

1.5849 = 0.10mol - [HA] / [HA]

1.5849 [HA] = 0.10mol - [HA]

2.5849[HA] = 0.10mol

[HA] = 0.0387 moles = H₂PO₄⁻ moles

That means moles of HPO₄²⁻ are 0.10mol - 0.0387moles = 0.0613 moles

The moles of NaOH needed to convert all H₃PO₄ in H₂PO₄⁻ are 0.10 moles

And moles needed to obtain 0.0613 moles of HPO₄²⁻ are 0.0613 moles

Total moles of NaOH are 0.1613moles * (1L / 1mol) = 0.1613L of 1.00M NaOH

Then, you need to dilute both solutions to 2.00L with distilled water.

4 0
3 years ago
C9H20 +<br> 02 - &gt; CO2 +<br> H2O<br><br> what is the balanced equation of this
san4es73 [151]

Hey there!

C₉H₂O + O₂ → CO₂ + H₂O

First let's balance the C.

There's 9 on the left and 1 on the right. So, let's add a coefficient of 9 in front of CO₂.

C₉H₂O + O₂ → 9CO₂ + H₂O

Next let's balance the H.

There's 2 on the left and 2 on the right. This means it's already balanced.

C₉H₂O + O₂ → 9CO₂ + H₂O

Lastly, let's balance the O.

There's 3 on the left and 19 on the right. So, let's add a coefficient of 9 in front of O₂.

C₉H₂O + 9O₂ → 9CO₂ + H₂O

This is our final balanced equation.

Hope this helps!

8 0
3 years ago
What is atom?????????????
kiruha [24]

Answer:

An atom is a particle of matter that uniquely defines achemical element. An atom consists of a central nucleus that is usually surrounded by one or more electrons. ... The nucleus is positively charged, and contains one or more relatively heavy particles known as protons and neutrons. A proton is positively charged.

6 0
2 years ago
Read 2 more answers
How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos
Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of H_3PO_4 = 54.219 g

Number of atoms of Mg = 7.179\times 10^{23}

Molar mass of H_3PO_4 = 98 g/mol

First we have to calculate the moles of H_3PO_4 and Mg.

\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}

\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol

and,

\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2

From the balanced reaction we conclude that

As, 3 mole of Mg react with 2 mole of H_3PO_4

So, 0.553 moles of Mg react with \frac{2}{3}\times 0.553=0.369 moles of H_3PO_4

From this we conclude that, H_3PO_4 is an excess reagent because the given moles are greater than the required moles and Mg is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of H_2

From the reaction, we conclude that

As, 3 mole of Mg react to give 3 mole of H_2

So, 0.553 mole of Mg react to give 0.553 mole of H_2

Now we have to calculate the volume of H_2  gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies 0.553\times 22.4=12.4L volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0
3 years ago
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