Determine the ph of a solution that results from the reaction of 75.0 ml 0.135 m hydrobromic acid with 95.0 ml of 0.115 m potassium hydroxide.

pH= 11.6727

Explanation: first we find the moles of HBr and KoH, but we have to firstly convert 75.0ml and 95.0ml to litre. I.e 75.0/1000 = 0.075L

95.0/1000 = 0.095L

Now we multiply the litres by their molarity to get the moles

0.075L × 0.0135mol/L = 0.010125mol HBr

0.095L * 0.115 mol/L = 0.010925mol KoH.

Subtracting mole of HBr from mole of KoH since KoH will remain after the reaction due to it been larger than HBr and KoH will dictate the pH

0.010925-0.010125= 0.0008 mol KoH left.

We find a new molarity of KoH

0.0008mol/ 0.075L+0.095L = 0.004706M KoH

To determine the POH

POH = -log(0.004706 M) = 2.3273

pH = total PH scale- POH

= 14- 2.3273 = 11.6727