Problem: Find the pH of the equivalence point(s) and the volume (mL) of 0.125 M HCl needed to reach the point(s) in titrations of(a) 65.5 mL of 0.234 M NH 3

FREE Expert Solution

Step 1

At equivalence: mol acid = mol base


mol NH3 = 65.5 mL ×10-3 L1 mL×0.234 mol NH31 L

mol NH3 = 0.015327 mol NH3


volume HCl =0.015327 mol NH3 × 1 mol HCl1 mol NH3×1 L HCl0.125 mol HCl×1 mL10-3 L

volume HCl = 123 mL



Step 2


Step 3


Step 4

Ka=[NH3][H3O+][NH4+]Ka = (x)(x)0.015327 - x


[NH4+]Ka=0.0153275.6818×10-10 >>> 500 → remove (-x)


5.6818 ×10-10 = x20.015327x2 = (5.6818 ×10-10)(0.015327)x2 = 8.7085×10-12


x = 2.9510x10-6 mol


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Problem Details

Find the pH of the equivalence point(s) and the volume (mL) of 0.125 M HCl needed to reach the point(s) in titrations of

(a) 65.5 mL of 0.234 M NH 3

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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Weak Base Strong Acid Titrations concept. You can view video lessons to learn Weak Base Strong Acid Titrations. Or if you need more Weak Base Strong Acid Titrations practice, you can also practice Weak Base Strong Acid Titrations practice problems.

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Based on our data, we think this problem is relevant for Professor Waddell's class at UC.

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Our data indicates that this problem or a close variation was asked in Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition. You can also practice Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition practice problems.