Answer :
When HNO3 (nitric acid) and RBOH (rubidium hydroxide) are mixed, they react in a neutralization reaction to form water and a salt, RbNO3.
This salt does not significantly affect the pH of the solution, which is determined by the remaining H+ and OH- ions in the solution. The balanced chemical equation for the reaction is:HNO3 + RBOH → RbNO3 + H2O
The molar ratios of HNO3 to RBOH in this reaction are 1:1. Therefore, to determine the number of moles of each compound in the solution, we can multiply the volume of each solution by its concentration, as shown below: moles of HNO3 = (166 mL)(1.55 M) = 257.3 m mol moles of RBOH = (133 mL)(1.37 M) = 182.2 m mol
Since the molar ratio of HNO3 to RBOH is 1:1, the reaction will use up all of the RBOH in the solution. The remaining H+ ions from the HNO3 will combine with the OH- ions from the RBOH to form water.
Therefore, the pH of the resulting solution will be determined by the concentration of H+ ions in the solution.
To calculate the concentration of H+ ions in the solution, we need to determine the number of moles of H+ ions in the solution. Since the molar ratio of HNO3 to H+ is also 1:1, we can assume that all of the HNO3 in the solution has dissociated into H+ and NO3- ions.
Therefore, the number of moles of H+ ions in the solution is equal to the number of moles of HNO3 in the solution, which is 257.3 m mol.
The volume of the resulting solution is the sum of the volumes of the two solutions, which is:166 mL + 133 mL = 299 mL
To calculate the concentration of H+ ions in the resulting solution, we can divide the number of moles of H+ ions by the volume of the solution and take the negative logarithm (base 10) of the result, as shown below: pH = -log[H+]where [H+] is the concentration of H+ ions in the solution.
Therefore: pH = -log(257.3 mmol / 299 mL) = 0.635Answer: The pH of the solution resulting from mixing 166 mL of a 1.55 solution of HNO3 with 133 mL of a 1.37M solution of RBOH is 0.635.
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