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Question

Calculate pH of the following solutions:
(i) 0.001MHNO3
(ii) 0.005MH2SO4
(iii) 0.01MKOH
(iv) 108MNaOH
(v) 0.0008MBa(OH)2

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Solution

Formula for pH=log[H+]
(i)0.001M HNO30.001M H+ pH=log[0.001] pH=3(ii)0.005M H2SO40.005×2M H+ pH=log[0.01] pH=2(iii)0.01M KOH0.01M OHp[OH]=log[0.01]pOH=2 pH=14pOH=142=12(iv)108M NaOH108M OHpOH=log[108] =8pH=148=6(v)0.0008M Ba(OH)22×0.0008M OH pOH=log[0.0016]=2.7959 pH=142.7959=11.2041

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