Chemistry, 23.07.2019 12:30 teesoprettyy

Calculate the ph of 0.375 l of a 0.18 m acetic acid-0.29 m sodium acetate buffer after the addition of 0.0070 mol of hbr. assume that the volume remains constant.

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Answer from: 19nnash

We are going to use H-H equation to calculate PH:

when:

PH = Pka + ㏒[A-]/[HA]

when Pka is the dissociation constant of the weak acid = 4.75

and we have [A-] (cocentration of conjugate base - acetate) = 0.29 M

and [ HA] (concentration of weak acid- acetic acid) = 0.18 M

when the reaction equation is :

CH3COOH + HBr ↔ CH3COO-Br + H2O

So, we need to calculate the moles of CH3COOH & CH3COO-Br

after adding HBr:

moles of acetic acid CH3COOH = molarity * volume

= 0.18 M * 0.375 L = 0.0675 moles

moles of acetate CH3COO- = molarity * volume

= 0.29 M * 0.375 L = 0.11 moles

so, the concentrations after adding HBr is:

[CH3COOH] =( moles CH3COOH - moles HBr)/total volume

= 0.0675 - 0.007 / 0.375L = 0.161 M

[CH3COO-] = (moles CH3COO- + moles HBr)/ total volume

=(0.11 mol + 0.007 mol)/0.375L = 0.312 M

by substitution in H-H equation:

∴ PH = 4.75 + ㏒(0.312/0.161)

= 5

Answer from: stricklandashley43

The pH of 0.375 L of a 0.18 M acetic acid-0.29 M sodium acetate buffer after the addition of 0.0070 mol of HBr : 4.9

Further explanation

A buffer solution is a solution that can maintain a good pH value due to the addition of a little acid or a little base or dilution.

The buffer solution can be acidic or basic

Acid buffer solution consists of weak acids and their salts.

$\displaystyle [H^+]=Ka\times\frac{mole\:weak\:acid}{mole\:salt\times valence}$

valence according to the amount of salt anion

Alkaline buffer solution consists of a weak base and its salt.

$\displaystyle [OH^-]=Kb\times\frac{mole\:weak\:base}{mole\:salt\times valence}$

valence according to the amount of salt cation

If the buffer solution is acidic then

a slight addition of acid (H⁺) will be balanced by the conjugate base

the addition of a small base (OH⁻) will be balanced by the weak acid

If the buffer solution is alkaline then

the addition of a little acid (H⁺) will be balanced weak base

the addition of a small base (OH⁻) will be balanced

Acetic acid-sodium acetate is an acid buffer solution

Reaction:

CH₃COOH + NaOH ↔ CH₃COO-Na + H₂O

mole of weak acid CH₃COOH =0.18 x 0.375=0.0675

mole of CH₃COONa =0.29 x 0.375=0.11

After addition of 0.0070 mol of HBr.

Reaction :

CH₃COONa + HBr ⇔ CH₃COOH + NaBr

0.11 0.007 0.0675

-0.007 -0.007 +0.007

====================================

0.103 0 0.0745

assumption pKa CH₃COOH = 4,76

[salt]=[CH₃COONa] = 0.103 : 0.375 = 0.275

[acid]=[CH₃COOH] = 0.0745 : 0.375 = 0.199

We can calculate [H+] :

$\displaystyle [H^+]=Ka\times\frac{mole\:weak\:acid}{mole\:salt\times valence}$

$\displaystyle pH=pKa+log\frac{[salt]}{[acid]}$

$\displaystyle pH=4.76+log\frac{0.275}{0.199}$

pH = 4.76 + 0.140 = 4.9

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pH scale

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solution has the lowest ph

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Keywords : buffer, acid, salt, pH

Calculate the ph of 0.375 l of a 0.18 m acetic acid-0.29 m sodium acetate buffer after the addition

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Answer from: Quest

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Calculate the ph of 0.375 l of a 0.18 m acetic acid-0.29 m sodium acetate buffer after the addition...