what will be the ph of 75ml 0.2M HCL +25 ml of 0.2 M of NAOH
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Answered by
1
Answer:
3
Explanation:
no.of mol of HCl = 75×0.2/1000= 0.015
no.of mol of NaOH = 25×0.2/1000 = 0.005
no.of mol of acid remaining =
= 0.010
[H+] = 0.010/(75+25)ml
= 0.010/0.1L= 0.0010mol/L
pH = -log(0.001)
= 3
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