A 4.36-g sample of an unknown alkali metal hydroxide is dissolved...

Question

A 4.36-g sample of an unknown alkali metal hydroxide is dissolved
in 100.0 mL of water. An acid–base indicator is added,
and the resulting solution is titrated with 2.50 M HCl(aq)
solution. The indicator changes color, signaling that the
equivalence point has been reached, after 17.0 mL of the hydrochloric
acid solution has been added. (a) What is the molar
mass of the metal hydroxide?

Accepted Answer

Hi everyone here we have a question telling us that a 108.64 g sample of an unknown Alkali metal hydroxide Is dissolved in ml of water. The solution is then tied traded with 161 ml of 4. molar nitric acid. To reach the equivalence point. And our goal is to calculate the molar mass of the middle hydroxide similarity equals moles over leaders. So to find moles, we simply multiply our polarity times our leaders And we are given ml so we need to change that to leaders. So we have 161 ml times one leader Over 10 to the 3rd male leaders. And our ml cancel out giving us 0.161 leaders. So now we can multiply our polarity times our leaders. So we have 4.5 Moeller nitric acid times zero point 161 L. And that equals 0.725 moles of nitric acid. And we're at the equivalence point. So that also equals 0.725 moles of our unknown alkali metal hydroxide, which I'm gonna do know as M. O. H. So then we have 108 . g and molar masses grams over moles. So we're going to divide that by our zero point 7- moles And that equals 150 graham's Permal and that is the molar mass of our unknown alkali metal hydroxide. Thank you for watching. Bye

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