Answer:
1.09
Step-by-step explanation:
Keep in mind that the volume of the solution changes during this titration, so to compute the amount of hydronium that is neutralized during this addition of base (in order to calculate the final pH of the solution), we must calculate the moles of all species in solution initially present. Because both NaOH and HCl ionize completely:initial mol OH−=mol NaOH=(0.0050 L)(0.10 molL)=0.00050 mol OH−initial mol H3O+=mol HCl=(0.050 L)(0.10 molL)=0.0050 mol H3O+The acid is in excess, so all of the OH− present will neutralize an equivalent amount of H3O+, forming water. Thus we simply subtract the moles of hydroxide from the moles of hydronium in solution to find the resultant moles of H3O+ after this neutralization:final mol H3O+= initial mol H3O+−initial mol OH−final mol H3O+=0.0050 mol−0.00050 mol=0.0045 mol H3O+We now calculate the total volume of the solution by adding the volumes of acid and base initially combined: 0.050 L+0.0050 L=0.055 LTo get [H3O+], we divide the final moles of hydronium by the final solution volume:[H3O+]=final mol H3O+ total volume=0.0045 mol0.055 L≈0.08181molLFinally, to find pH:pH=−log[H3O+]=−log(0.08181)=1.09Since the hydronium concentration is only precise to two significant figures, the logarithm should be rounded to two decimal places.