# 11 chemistry ionic equilibrium

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How many grams of solid NaOH must be added to 100ml of buffer solution which is 0.1M each w.r.t acid HA and salt Na+A- to make the pH of the solution 5.5

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0.2 g NaOH

Explanation:

You know that the hydroxide anions delivered to the solution by the sodium hydroxide will react with the weak acid to produce the conjugate base in 1:1 mole ratios.

HA(aq)+OH−(aq)→A−(aq)+H2O(l)

The number of moles of weak acid and of conjugate base present in the buffer before the addition of the strong base is given by

moles HA=moles A−=0.1 moles103  mL⋅100mL

moles HA = moles A−=0.01 moles

Now, if you take x to be the number of moles of sodium hydroxide added to the buffer, you can say that after the reaction is complete, the resulting solution will contain

(0.01−x)  moles HA

The reaction will consume x moles of the weak acid.

(0.01+x)  moles A−

The reaction will produce x moles of the conjugate base.

As you know, the pH of a weak acid-conjugate base buffer can be calculated using the Henderson - Hasselbalch equation.

pH=pKa+log([A−][HA])

Assuming that the volume of the buffer does not change upon the addition of the strong base, you can say that the volume of the solution after the strong base is added is equal to 100 mL.

That said, the fact that the volume is the same for both the weak acid and the conjugate base allows you to treat the concentrations of the two chemical species and the number of moles interchangeably.

You can thus say that after the strong base is added, the pH of the solution will be equal to

pH=pKa+log[(0.01+x)moles(0.01−x)moles]

Plug in your values to find

5.5=5+log(0.01+x0.01−x)

This will be equivalent to

log(0.01+x0.01−x)=0.5

To find the value of x, rewrite this as

10log(0.01+x0.01−x)=100.5

You will ned up with

0.01+x0.01−x=3.16

0.01+x=0.0316−3.16x

So

x=0.0316−0.011+3.16=0.00519

Since x represents the number of moles of sodium hydroxide added to the buffer, you can use the molar mass of the compound to convert this to grams.

0.0519moles NaOH⋅39.997 g1mole NaOH=0.2 g−−−−

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