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2 years ago

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You have a 1.8 M solution of HCI. You need to make a 0.70 M solution with a volume of 550 mL. How many mL of the stock solution

do you need to dilute to get your new concentration?

1 answer:

Arte-miy333 [17]2 years ago

5 0

The volume of the stock solution needed is 213.88 mL to get new concentration.

<h3>What is molarity?</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution.

Determination of the volume of stock solution.

Volume of diluted solution (V₂) =550 mL

Molarity of diluted solution (M₂) =0.70 M

Molarity of stock solution (M₁) = 1.8 M

Volume of stock solution needed (V₁) =?

M₁V₁ = M₂V₂

1.8 M × V₁ = 0.70 M × 550 mL

V₁ = 213.88 mL

Thus, the volume of the stock solution needed is 213.88 mL.

Learn more about the molarity here:

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You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.8

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Complete Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?

Answer:

The pK_a value is $pK_a =7.82$

Explanation:

From the question we are told

The volume of base is $V_B = 26.mL = 0.0260L$

The pH of solution is $pH = 7.82$

The concentration of the acid is $C_A = 0.1M$

From the pH we can see that the titration is between a strong base and a weak acid

Let assume that the the volume of acid is $V_A = 18mL= 0.018L$

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$C_B = \frac{C_AV_A}{C_B}$

Substituting value

$C_B = \frac{0.1 * 0.01800}{0.0260}$

$C_B= 0.0692M$

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The chemical equation of the reaction is

$HA_{(aq)} + OH^-_{(aq)} --------> A^{+}_{(aq)} + H_2 O_{(l)}$

Now before the reaction the number of mole of base is

$No \ of \ moles[N_B] = C_B * V_B$

Substituting value

$N_B = 0.01300 * 0.0692$

$= 0.0009 \ moles$

Now before the reaction the number of mole of acid is

$No \ of \ moles = C_B * V_B$

Substituting value

$N_A = 0.01800 *0.1$

$= 0.001800 \ moles$

Now after the reaction the number of moles of base is zero i.e has been used up

this mathematically represented as

$N_B ' = N_B - N_B = 0$

The number of moles of acid is

$N_A ' = N_A - N_B$

$= 0.0009\ moles$

The pH of this reaction can be mathematically represented as

$pH = pK_a + log \frac{[base]}{[acid]}$

Substituting values

$7.82 = pK_a +log \frac{0.0009}{0.0009}$

$pK_a =7.82$

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