Volume of 0.150 m naoh solution required to neutralize 25ml of a 0.055 m hcl solution?
equation of the reaction: naoh(aq) + hcl(aq) > h2o(l) + nacl(aq)

9.17 mL

In this question we are given;

NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)

We are required to determine the volume of the alkali, NaOH used;

Molarity = Number of moles ÷ Volume

Thus, rearranging the formula;

Moles = Molarity × volume

          = 0.055 M × 0.025 L

          = 0.001375 moles

Therefore;

Hence; Moles of NaOH = 0.001375 moles

Molarity = Moles ÷ Volume

Volume = Moles ÷ Molarity

Therefore;

Volume of NaOH = 0.001375 moles ÷ 0.150 M

                             = 0.00917 L, but, 1 L = 1000 mL

                             = 9.17 mL

Therefore, the volume of NaOH required is 9.17 mL

it is a good conductor

explanation:

i'm assuming that the correct answer is indeed the first one/ "a"..