calculate the pH value of 0.005 molesof h2so4
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H2SO4 will dissociate as:
H2SO4→2H^+ +SO4²^-
0.01Moles of H2SO4 will give 2×0.01=0.02moles of H^+
KOH→K^+ +OH^-
0.01 Moles of KOH will give 0.01 moles of H^+ from H2SO4,0.01moles of H^+ will be neutralized by 0.01 moles of OH^- remaining in 2L water will be 0.02-0.01 concentration of H^+=0.01moles/2L=O.OO5moles/L=5×10^-3 moles/L
pH=-㏒(H+)
pH=-㏒(5×10^-3)
pH=3-㏒5
pH=3-0.6989
pH=2.3011
H2SO4→2H^+ +SO4²^-
0.01Moles of H2SO4 will give 2×0.01=0.02moles of H^+
KOH→K^+ +OH^-
0.01 Moles of KOH will give 0.01 moles of H^+ from H2SO4,0.01moles of H^+ will be neutralized by 0.01 moles of OH^- remaining in 2L water will be 0.02-0.01 concentration of H^+=0.01moles/2L=O.OO5moles/L=5×10^-3 moles/L
pH=-㏒(H+)
pH=-㏒(5×10^-3)
pH=3-㏒5
pH=3-0.6989
pH=2.3011
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