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3 years ago

12

. What happens the pH of an acid solution as a base is added?

1 answer:

Dmitrij [34]3 years ago

3 0

As a base is added to an acidic solution, the H+ ions in solution that make it acidic are slowly neutralized into water (via OH-, the base). As these ions are converted into water the concentration of them decreases, so the pH decreases, as they are directly related.

Hope this helps!

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Suppose you need to prepare 136.9 mL of a 0.315 M aqueous solution of NaCl.

Varvara68 [4.7K]

Answer:

2.52 g NaCl

Explanation:

(Step 1)

To find the mass, you first need to find the moles NaCl. This value can be found using the molarity ratio:

Molarity = moles / volume (L)

After you convert mL to L, you can plug the given values into the equation and simplify to find moles.

136.9 mL / 1,000 = 0.1369 L

Molarity = moles / volume

0.315 M = moles / 0.1369 L

0.0431 = moles

(Step 2)

Now, you can use the molar mass to convert moles to grams.

Molar Mass (NaCl): 22.990 g/mol + 35.453 g/mol

Molar Mass (NaCl): 58.443 g/mol

0.0431 moles NaCl 58.443 g
------------------------------ x ------------------- = 2.52 g NaCl
1 mole

4 0

2 years ago

atroni [7]

The last one would be false

3 0

4 years ago

Read 2 more answers

Compute the product of P.V.<br><br> The Awnser is 51.3

Soloha48 [4]

The answer is 51.3 ...

6 0

3 years ago

2 HCl + Na2SO4 > 2 NaCl + H2SO4

Cerrena [4.2K]

Answer:

15

Explanation:

8 0

3 years ago

People take antacids, such as milk of magnesia, to reduce the discomfort of acid stomach or heartburn. The recommended dose of m

Llana [10]

Answer:

$V_{HCl}=0.208L=208mL$

Explanation:

Hello,

In this case, since the chemical reaction is:

$2HCl+Mg(OH)_2\rightarrow MgCl_2+2H_2O$

We can see that hydrochloric acid and magnesium hydroxide are in a 2:1 mole ratio, which means that the neutralization point, we can write:

$n_{HCl}=2*n_{Mg(OH)_2}$

In such a way, the moles of magnesium hydroxide (molar mass 58.3 g/mol) in 500 mg are:

$n_{Mg(OH)_2}=500mg*\frac{1g}{1000mg}*\frac{1mol}{58.3g} =0.00858mol$

Next, since the pH of hydrochloric acid is 1.25, the concentration of H⁺ as well as the acid (strong acid) is:

$[H^+]=[HCl]=10^{-pH}=10^{-1.25}=0.0562M$

Then, since the concentration and the volume define the moles, we can write:

$[HCl]*V_{HCl}=2*n_{Mg(OH)_2}$

Therefore, the neutralized volume turns out:

$V_{HCl}=\frac{2*0.00858mol}{0.0562\frac{mol}{L} }\\ \\V_{HCl}=0.208L=208mL$

Best regards.

3 0

3 years ago