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Determine the concentration of strontium hydroxide if 45 mL of the base are neutralized by 32 mL of 0.30M chloric acid. Show all work and round answers to the correct number of significant figures.

Sagot :

Answer

The concentration of strontium hydroxide = 0.11 M

Explanation

Given:

Volume of strontium hydroxide, Vb = 45 mL

Volume of chloric acid, Va = 32 mL

Concentration of chloric acid, Ca = 0.30 M

What to find:

The concentration of strontium hydroxide, Cb.

Step-by-step solution:

Step 1: Write a balanced equation for the neutralization reaction.

[tex]Sr(OH)_2+2HClO_3\rightarrow Sr(ClO_3)_2+2H_2O[/tex]

Mole ratio from the equation is 1:2, that is nb = 1 and na = 2

Step 2: Determine the concentration of strontium hydroxide, Cb.

Using the formula below;

CaVa/na =CbVb/nb

So Cb = (Ca x Va x nb)/(Vb x na)

Plugging the values of the given parameters into the equation, we have

Cb = (0.30M x 32mL x 1)/(45mL x 2)

Cb = 9.6 M/90 m

Cb = 0.11 M

The concentration of strontium hydroxide = 0.11 M

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