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Grace [21]
3 years ago
11

In a titration how much 0.50 M HCl is needed to neutralize 1 liter of a 0.75 M solution of NaOH?

Chemistry
1 answer:
Ymorist [56]3 years ago
6 0

Answer:

1.5 L

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

HCl + NaOH —> NaCl + H2O

From the balanced equation above, The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question. This includes the following:

Molarity of acid (Ma) = 0.5M

Volume of acid (Va) =..?

Volume of base (Vb) = 1L

Molarity of base (Mb) = 0.75M

Step 3:

Determination of the volume of the acid needed for the reaction.

Using the formula:

MaVa/MbVb = nA/nB

The volume of the acid needed for the reaction can be obtained as follow:

0.5 x Va / 0.75 x 1 = 1

Cross multiply

0.5 x Va = 0.75 x 1

Divide both side by 0.5

Va = 0.75 /0.5

Va = 1.5 L

Therefore, the volume of the acid, HCl needed for the reaction is 1.5L

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Adopting the number of avogrado 6.02 * 10²³ / mol 
<span>Sodium chloride (table salt)</span> Molar Mass = 58.44 g / mol 
We will first have to find the number of moles in 35 grams of the element, like this: 
 
1 mol ----------------- 58.44 g 
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58.44 * x = 35 * 1 
58.44x = 35 
X = \frac{35}{58.44}
X = 0.598904...
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Now we will find how many atoms there are in 0.60 mol of this element, like this: 
 
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X = 0.60 * 6.02 * 10²³ 
\boxed{\boxed{x \approx 3.612 * 10^{23}\:atoms}}\end{array}}\qquad\quad\checkmark
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