Solution:
((25 mL x 0.016 M) + (35 mL x 0.041 M)) / (25 mL + 35 mL)
= 0.0305 M HCl
Here HCl is a strong acid, so [H+] = [HCl]:
pH = - log 0.0305
= 1.5
Option C is correct
17. What is the pH of the final solution when 25 ml.of 0.016 MHCI has been...
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For this question, we have a 17 ml solution of 1.7M ammonia, Kb=1.8x105 a) What is the initial pH? [number 1] b) What is the pH when 40 mL of 0.17 M HCl has been added? [number2] c) Where was the equivalence point volume? (number3] d) What is the pH when 8 mL of HCl has been added? [number4]
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