Answer:
9.15
Explanation:
To solve this problem we'll use Henderson-Hasselbach's equation:
For this problem:
Now we calculate [NH₄Cl], first by converting 50.0 g of NH₄Cl into moles using its molar mass:
Meaning that [NH₄Cl] = 0.935 mol / 1.00 L = 0.935 M
Finally we calculate the pH:
The pH can be defined as the hydrogen ion concentration in the solution. The pH of ammonium chloride solution in ammonia is 9.16.
The molarity is the moles of a compound in the liter of solution. The molarity of 50 grams ammonium chloride in a liter of solution is given as:
[tex]\rm Molaity=\dfrac{mass}{molar\;mass\;\times\;volume(L)} \\\\NH_4Cl\;Molarity=\dfrac{50\;g}{53.491\;g/mol\;\times\;1\;L} \\\\NH_4Cl\;Molarity=0.935\;M[/tex]
The ammonia in the solution contributes to the pH from the ammonium chloride salt. The pH can be given as:
[tex]\rm pH=pKa+log\dfrac{acid}{salt}[/tex]
The pKa of ammonia is 9.25. The concentration of ammonia is given as 0.75 M.
The pH of the solution can be given as:
[tex]\rm pH=9.25+log\dfrac{0.75\;M}{0.935}\\ pH=9.25+(-0.09)\\pH=9.16[/tex]
The pH of the solution is 9.16.
Learn more about pH, here:
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