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

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the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.