Question

In the simulation, open the Custom mode. The beaker will be filled to the 0.50 L mark with a neutral solution. Set the pH to 7.30 by using the green arrows adjacent to the pH value indicated on the probe in the solution. Once you adjust the pH, note the corresponding OH− ion concentration in M as given in the graphic on the left side of the simulation. Make sure to select the option "Concentration (mol/L)" above the graphic. Select on the Logarithmic scale below the graphic.

It says to find the pOH of the solution. We haven't really gone over this in class and the teacher says that this won't even really help of study for this class but we need the homework credit...

Answer 1

Okay, I can help you find out the hydroxide(OH-) ion concentration and pOH of the solution based on the information you have provided in the question.

It may be noted that pH = - log [H+] and pOH = - log [OH-] where square brackets indicate molar concentrations.

Initially, the beaker is filled to 0.5 L mark with neutral solution. Although the temperature in this question is not mentioned, but considering 298 K which is taken as 'room temperature', the pH of a neutral solution is 7.0. At this pH, a given solution has equal concentrations of H+ and OH- ion, each being equal to 10^-7 M. At pH lower than 7, the solution has excess of H+ ions and at pH higher than 7 the solution has excess of OH- ions.

Therefore for all aqueous solutions at 298 K, the product of H+ and OH- ion concentrations is 10^-14. It is called ionic product of water, Kw.

When the pH of the solution is adjusted to 7.3, it becomes alkaline i.e., has excess of hydroxide ion concentration.

hope this helps:)

Now, = x OR [+] [H] = 10-14 Taking logarithm on both sides log (4+] & log côt] =– 14 (log 10 - log [47] - log côn3 = 14 pH & pOH = 14 using ph=7.3, we find pOH = 14-7.3 = POH = 6.7 Also pOH = -log [on] = 6.7 OR log Lôt] = -6.7 OR Coh ] = antilog (-6.7) Con = 2x107 m)