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. (b) What is the identity of the alkali metal cation: Li⁺, Na⁺, K⁺, Rb⁺, or Cs⁺?

Video transcript

Hi everyone here, we have a question telling us that a 108.64 g sample Of an unknown Alkali metal hydroxide is dissolved in 150 ml of water. The solution is then tie traded with 161 ml of 4.5 molar nitric acid to reach the equivalence point. And our goal here is to identify the alkali metal. So malaria T equals moles over leaders. So to find the moles, we just multiply our polarity times our leaders And it gives us mill leaders. So we want to change that to leaders. So we have 161 ml times one liter Over 10 to the 3rd mill leaders. So our middle leaders are going to cancel out here and that's going to equal 0.161 L. So now we can take our 4. moller nitric acid And multiply it by our 0.161 leaders. And that gives us 0.725 moles of nitric acid. And because we are at the equivalence point, that also equals 0.725 moles of our unknown alkali metal hydroxide, which I'm going to know as M O. H. Now to find the molar mass. Our molar mass is grams divided by moles. So we're going to take our 108 .64 g Divided by R 0.725 moles. And that equals .84 g per mole. And our molar mass of our middle is going to equal the molar mass of our middle hydroxide minus the molar mass of hydroxide. So the molar mass of our medal Is going to equal R 149 0. grams for mole minus the mass of oxygen, which is 16 Plus the mass of hydrogen, which is 1.01 And that equals 132.84 graham's Permal. And if we look on the periodic table, the element that is closest to that is cesium. So our unknown medal here is cesium. Thank you for watching. Bye!