Neutralization involves an acid donating a Hydrogen (H+) to a base (in this case, NaOH) to form a salt and water. We need to find the moles of H+ being neutralized by NaOH:
moles of H+ from HCl:
295mL(1L/1000mL)(0.173mol HCl/L)(1mol H+/mol HCl)=0.05104 mol H+ being neutralized by NaOH
moles of H+ from H2SO4:
295mL(1L/1000mL)(0.346mol H2SO4/L)(2mol H+/mol H2SO4)=0.2041 mol H+ being neutralized by NaOH
total moles of H+ neutralized by NaOH:
0.05104 mol + 0.2041 mol = 0.25514 moles of H+
Now you use the total moles of H+ neutralized to determine the volume of NaOH that was needed (OH- and H+ react in a 1:1 molar ratio to form water by the equation H+(aq) + OH-(aq)—> H2O(l)):
0.25514mol H+(1mol NaOH/1mol H+)(1L NaOH solution/0.260mol NaOH)(1000mL/L) = 981 mL NaOH solution
So the answer is 981 mL NaOH solution!
