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2 years ago

9

Determine the mass in grams of 0.0489 mol cobalt

1 answer:

Alexus [3.1K]2 years ago

6 0

Answer:

2.88 g

Explanation:

Multiply the given number of moles by the molar mass to get the grams.

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The concentration of a certain sodium hydroxide solution was determined by using the solution to titrate a sample of potassium h

7nadin3 [17]

Answer:

0.022 M

Explanation:

The molarity is the number of mol solute divided by the volume of solution in liters. Since we are told there is only one acidic hydrogen in KHP the reaction is a 1 to 1 with sodium hydroxide, We then calculate the moles of sodium hydroxide required to react with the moles of KHP and proceed to calculate the molarity of the NaOH solution:

mol KHP: 0.1082 g x 1 mol/204.22 g = 0.00053 mol KHP

0.00053 mol KHP x 1 mol Naoh/1 mol KHP = 0.00053 mol NaOH

Molarity of the solution : 0.00053 mol NaOH / 0.02346 L = 0.022 M

note: The volume of sodium hydroxide needed to be converted to liters by definition of molarity.

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3 years ago

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Elodia [21]

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Answer:

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3 years ago

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Vlada [557]

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3 years ago

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Phosphoric acid is a triprotic acid ( K a1 = 6.9 × 10 − 3 Ka1=6.9×10−3, K a2 = 6.2 × 10 − 8 Ka2=6.2×10−8, and K a3 = 4.8 × 10 −

irina1246 [14]

<u>Answer:</u> To calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

<u>Explanation:</u>

Phosphoric acid is a triprotic acid and it will undergo three dissociation reaction each having their respective dissociation constants.

The chemical equation for the first dissociation reaction follows:

$H_3PO_4\rightleftharpoons H_2PO_4^-+H^+;K_a1=6.9\times 10^{-3}$

The chemical equation for the second dissociation reaction follows:

$H_2PO_4^-\rightleftharpoons HPO_4^{2-}+H^+;K_a2=6.2\times 10^{-8}$

The chemical equation for the third dissociation reaction follows:

$HPO_4^{2-}\rightleftharpoons PO_4^{3-}+H^+;K_a3=4.8\times 10^{-13}$

To form a buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a$ of second dissociation process

To calculate the $pK_a$ , we use the equation:

$pK_a=-\log (K_a)\\\\pK_a=-\log(6.2\times 10^{-8})\\\\pK_a=7.21$

To calculate the pH of buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a2+\log(\frac{[\text{conjugate base}]}{[\text{weak acid}]})$

$pH=pK_a2+\log(\frac{[HPO_4^{2-}]}{[H_2PO_4^-]})$

We are given:

$pK_a2$ = negative logarithm of second acid dissociation constant of phosphoric acid = 7.21

$[HPO_4^{2-}]$ = concentration of conjugate base

$[H_2PO_4^{-}]$ = concentration of weak acid

Hence, to calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

3 0

4 years ago