Answer :
The pH of a 5g HF solution in 100 ml of water is -0.398 (acidic) and the pH of a 5g RbOH solution in 100 ml of water is 15.31 (basic). It was assumed that the HF and RbOH completely ionize in solution, despite HF being a weak acid.
This question is about calculating the pH of the solutions of two different substances dissolved in water. The pH of a solution can be calculated using the formula: pH = -log10[H+], where [H+] is the concentration of hydronium ions in the solution.
Firstly, the molar mass of HF (Hydrofluoric Acid) is approximately 20 g/mol. Therefore, when 5 g of HF is dissolved in water, it forms approximately 0.25 moles of HF. The volume of the solution is 100 ml or 0.1 L, so the concentration of HF is 0.25 moles/0.1 L = 2.5 M (moles per litre).
Hydrofluoric acid is a weak acid and does not fully dissociate into its ions in solution. However, for the purpose of this exercise, let's assume the ionization is complete and [H+] = 2.5 M. The pH of the solution is then -log10(2.5) = -0.398, which is acidic.
Secondly, for 5.00 g of RbOH (Rubidium Hydroxide) dissolved in 100 ml of water, we first calculate the moles of RbOH based on its molar mass of approximately 102.5 g/mol. This gives us an approximate molarity of 0.049 M. As RbOH is a strong base, it dissociates completely in water to form Rb+ and OH- ions.
To find the pH of the solution, we must first find the pOH, using the formula pOH = -log10[OH-]. From this, we can calculate the pH using the formula pH = 14 - pOH.
Therefore, the pH of the solution is approximately 14 - (-1.31) = 15.31, indicating a very basic solution.
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