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3 years ago

What is the mass, in grams, of 1.58 mol of water, h2o?

1 answer:

3 0

Answer is: mass of 1,58 mol of water is 28,44 grams.
n(H₂O) = 1,58 mol.
M(H₂O) = 2 · 2 g/mol + 16 g/mol.
M(H₂O) = 18 g/mol.
m(H₂O) = n(H₂O) · M(H₂O).
m(H₂O) = 1,58 mol · 18 g/mol.
m(H₂O) = 28,44 g.
n - amount of substance.
M - molar mass.

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3 years ago

For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ

Komok [63]

Answer:

Check the explanation

Explanation:

Answer – Given, $H_3PO_4$ acid and there are three Ka values

$K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}$

The transformation of $H_2PO_4- (aq) to HPO_4^2-(aq)$ is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

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First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

= 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

= 0.225 moles

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Now we need to calculate the pKa2

pKa2 = -log Ka

= -log 6.2x10-8

= 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

= 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

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3 years ago

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aleksandrvk [35]

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