Answer to Question #96268 in General Chemistry for Boye Zephaniah

Question #96268
How many mL of 0.1M HNO3 are required to neutralize exactly half of the OH ion present in 520 mL of 0.18N KOH
Expert's answer

n KOH =n OH = 0.52*0.18= 0.0936 mol

V HNO3 = n OH / C HNO3 = 0.0936/ 0.1= 0.936L = 936mL

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