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VMariaS [17]
3 years ago
15

The ph of a solution prepared by mixing 50.0 ml of 0.125 m naoh and 40.0 ml of 0.125 m hno3 is ________.

Chemistry
1 answer:
CaHeK987 [17]3 years ago
4 0
Both the acid and the base are known as strong substances which means they totally dissociate into ions in aqueous solution. Mixing the two a salt would be form which is NaNO3. The unreacted substances will be the one contributing to the pH of the solution. 

0.125M HNO3 x 0.04L = 0.005 mol HNO3
<span>0.125M NaOH x 0.05L = 0.00625 mol NaOH
</span>
As we can see, all of the HNO3 would react with NaOH so the remaining OH from the unreacted NaOH would be responsible for the pH.

0.00625 - 0.005 / ( 0.0500L + 0.0400L ) = 0.0139 M 

<span>pOH = -log( OH- ) </span>
<span>pOH = 1.86 </span>
<span>pH = 14.00 - pOH </span>
<span>pH = 12.14</span>
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0.595 M

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0.0194 = x/x + 55.6

0.0194(x + 55.6) = x

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If

xA= mole fraction of solvent

xB= mole fraction of solute

nA= number of moles of solvent

nB = number of moles of solute

MA= molar mass of solvent

MB = molar mass of solute

d= density of solution

Molarity = xBd × 1000/xAMA ×xBMB

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