How many mL of 0.230 m Ca(OH) 2 is needed toneutralize 87.2 of 0.4151 m HClO 4 ?

Question

How many mL of 0.230 m Ca(OH)2 is needed toneutralize 87.2 of 0.4151 m HClO4? this is what i came up with..... 1mol H+ / 1mol HCLO4 x 0 .4151 HClO4 mol/L x 0.0872 L = 0.150 mol Ca(OH)2 / L   x VOL(L)                                                                                                                x 2(OH)- / 1mol Ca(OH)2 Problem: the L signs seem to cancel out and i'm leftwith 1mol H+ x .0362 =   0.3(OH)-                           I don't know what to do next. How many mL of 0.230 m Ca(OH)2 is needed toneutralize 87.2 of 0.4151 m HClO4? this is what i came up with..... 1mol H+ / 1mol HCLO4 x 0 .4151 HClO4 mol/L x 0.0872 L = 0.150 mol Ca(OH)2 / L   x VOL(L)                                                                                                                x 2(OH)- / 1mol Ca(OH)2 Problem: the L signs seem to cancel out and i'm leftwith 1mol H+ x .0362 =   0.3(OH)-                           I don't know what to do next. Problem: the L signs seem to cancel out and i'm leftwith 1mol H+ x .0362 =   0.3(OH)-                           I don't know what to do next.

Explanation / Answer

first lets figure out how many H+ ions there are in moles by multiplaying .0872 L by molarity to get moles so. .0872*.4151=0.0361967 so that means we have 2OH- per H+ from the calcium hydroxide so we only need half as many moles of Ca(OH)2 as H+ so 0.0180984 moles of Calcuim hydroxide so... we get 0.0180984 moles *molarity to get liters neceary so 0.0180984 moles*(1L/.23 moles) this cancels out moles and leaves uswith liters so 0.0786885 Liters and it asks us in mL so 78.7

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