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
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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


100% (351 ratings)
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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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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