How many milliliters of .128M H2SO4 are needed to neutralize .210g of NaOH?
Well for starters Sulfuric Acid, H2SO4, is a strong acid and Sodium Hydroxide, NaOH, is a strong base. So they will neutralize each other with the same effect. Now 0.128 M means taht there are 0.128 moles of H+ ions for every liter of solution. It is now necessary to break up the NaOH into just OH- so that you can figure out how much acidic solution you need. Since Sodium’s molecular weight is 23 g, Oxygen is 16 g, and Hydrogen is 1 g, you have to figure out the respective ratio to apply to the mass of NaOH. Therefore, (16+1)/(16+1+23) or 17/40 which is 0.425 is the mass fraction of hydroxide in NaOH. This gives you 0.08925 g of OH- to neutralize. The amount of moles in this is 0.08925/17 = 0.00525 moles of OH-. Lastly you have to divide this number by the molarity of the acidic solution to find out just how much acidic solution you need to completely neutralize the base. So, 0.00525/0.128 = 0.041016 L = 41.016 mL of H2SO4 to neutralize the 0.210 grams of NaOH. Hope I was of some help
