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3 years ago

what volume of a 0.149 m potassium hydroxide solution is required to neutralize 17.0 ml of a 0.112 m hydrobromic acid solution?

Chemistry

1 answer:

IgorLugansk [536]3 years ago

5 0

Answer: 12.78ml

Explanation:

Given that:

Volume of KOH Vb = ?

Concentration of KOH Cb = 0.149 m

Volume of HBr Va = 17.0 ml

Concentration of HBr Ca = 0.112 m

The equation is as follows

HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)

and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)

Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb

(0.112 x 17.0)/(0.149 x Vb) = 1/1

(1.904)/(0.149Vb) = 1/1

cross multiply

1.904 x 1 = 0.149Vb x 1

1.904 = 0.149Vb

divide both sides by 0.149

1.904/0.149 = 0.149Vb/0.149

12.78ml = Vb

Thus, 12.78 ml of potassium hydroxide solution is required.

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choli [55]

Answer:

$\boxed {\boxed {\sf 9.15*10^{24} \ atoms \ K}}$

Explanation:

To convert atoms to moles, Avogadro's Number must be used: 6.022*10²³.

This tells us the amount of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case it is the atoms of potassium. We can create a ratio.

$\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }$

Multiply by the given number of moles: 15.2

$15.2 \ mol \ K *\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }$

The moles of potassium cancel.

$15.2 *\frac {6.022*10^{23} \ atoms \ K }{ 1 }$

The denominator of 1 can be ignored.

$15.2 * {6.022*10^{23} \ atoms \ K }{$

Multiply.

$9.15344*10^{24} \ atoms \ K$

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated that is the hundredth place. The 3 in the thousandth place tells us to leave 5.

$9.15*10^{24} \ atoms \ K$

In 15.2 moles of potassium, there are 9.15*10²⁴ atoms of potassium.

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