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. Part B What is the identity of the alkali metal cation: Li+, Na+, K+, Rb+, or Cs+? Li+ Na+ K+ Rb+ Cs+

Part A What is the molar mass of the metal hydroxide?

Answer 1

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Alkali metal & Lit, Nat, kt, R6 or cst - Mt It posess (11) charge so, if forms M(OH) nanohydroxide The reaction ww mot and he is MOH (ag) + Hlllagg Melrag) + H₂0 (8) | I mot not there we use molarity formula; molarity - no. of moles (moles) volume of solution (L) Man In = mxv] til Molarity = 2.50M = 2.5 moth; volume = 120mL = oogol n= (3.5cmol/L) (06474) -0.0425 med MOH) from equation, 1 mol mot reacts with I mot Hel(lemot un 0.0425 mot Mot dearts with 0.0425 ml He. Therefore, 0.0425 mol of metal hydroxide contains 4.36g. # no. of mole - mass of colute (mg) molor mass of solute (g/mol) molar mass = (n) no. of notes (mol) mals of soluté (9) mass of solute t o no. of mole (mol) - 4-369 0.0425 mol = 102.58 g/mol Now, calculate the molar mass of melal (M)+(0)+1(A) = 102.58 g/mol 1(M) + (169/met) +(1 g/mol) = 102.58 g/mol 1(M) + (16 + 1 g/mol = 102. Sagland 1 - 102. seglme – 17 g/mol IM = 85.58 g/mol The molar mass of the alkali metal is 85.58 gloads This is the molar mass of Rubidium(R) out of all alkalimetels, So, onetal hydroxide - RbOH Rubidas hydroxide