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mart [117]
3 years ago
11

Why was it necessary to make sure that some solid was present in the main solution before taking the samples to measure Ksp?

Chemistry
1 answer:
padilas [110]3 years ago
7 0

The question is incomplete, the complete question is;

Why was it necessary to make sure that some solid was present in the main solution before taking the samples to measure Ksp? Select the option that best explains why. Choose... A. To make sure no more sodium borate would dissolve in solution. B. To ensure the dissolution process was at equilibrium. C. To make sure the solution was saturated with sodium and borate ions. D. All of the above

Answer:

B. To ensure the dissolution process was at equilibrium.

Explanation:

The solubility product is a term used in chemistry to describe the equilibrium between the dissolved, dissociated and undissolved solute of a relatively low solubility ionic solid.

For an ionic solid MX, the solubility product is given as ;

MX(s) ----> M^n+(aq) + X^n-(aq)

If Ksp indeed scribes an equilibrium process for dissolution, it then implies that some undissolved solute must be present before samples are taken to measure the Ksp of a sample. This ensures equilibrium between dissolved and undissolved solute.

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Answer:

\boxed {\boxed {\sf 0.495 \ mol}}

Explanation:

We are given a number of particles and asked to convert to moles.

<h3>1. Convert Particles to Moles </h3>

1 mole of any substance contains the same number of particles (atoms, molecules, formula units) : 6.022 *10²³ or Avogadro's Number. For this question, the particles are not specified.

So, we know that 1 mole of this substance contains 6.022 *10²³ particles. Let's set up a ratio.

\frac { 1 \ mol }{6.022*10^{23 } \ particles}}

We are converting 2.98*10²³ particles to moles, so we multiply the ratio by that value.

2.98*10^{23} \ particles *\frac { 1 \ mol }{6.022*10^{23 } \ particles}}

The units of particles cancel.

2.98*10^{23}  *\frac { 1 \ mol }{6.022*10^{23 } }}

\frac { 2.98*10^{23}}{6.022*10^{23 } }}  \ mol

0.4948522086 \ mol

<h3>2. Round</h3>

The original measurement of particles (2.98*10²³) has 3 significant figures, so our answer must have the same.

For the number we found, 3 sig figs is the thousandth place.

The 8 in the ten-thousandth place (0.4948522086) tells us to round the 4 up to a 5 in the thousandth place.

0.495 \ mol

2.98*10²³ particles are equal to approximately <u>0.495 moles.</u>

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