100mL of solution contain 0.005 moles of Ba(OH)2 ∴1L(1000mL) of solution contain =10×0.005‌ mole of ‌Ba(OH)2 Concentration of Ba(OH)2 (i.e., moles in litres.) =0.05M EachBa(OH)2 give 2OH−ions. Thus, moles of OH−(per L.) =2×0.05=0.1 (iii) ∵pOH=−log[OH−] pOH=−log[0.1]=−log‌10−1 pOH=log‌10=1