Question

6)(9) Consider the titration of 30.00 mL 0.300 M HCl(aq) a SA with 0.450 M NaOH(aq) a SB. a) Write an equation for the reaction between HCl and NaOH. b) What volume of NaOH must be added to reach the equivalence point in this titration? c) What is the pll at the equivalence point in this titration (no calculation required)?

Answer 1

a) HCl + NaOH gives NaCl + H2O( This is a neutralization reation).

b) We can use the formula MV = constant where M is the molarity and V is the volume of the solution so we can now write M1V1 = M2V2 where M1 is the molarity of the HCl and V1 is the volume of HCl used. Similarly, M2 is the molarity of NaOH used and V2 is the volume of NaOH. When mixing the two solutions the number of moles remain constant. hence the product of MV is constant as it is nothing but number of moles.( M= n/V ; where M is the molarity, V is the volume and n is the number of moles). So now we can apply M1V1=M2V2 ,here let us suppose the volume of NaOH as x , so on calculating , 30 * 0.30 = 0.450* x, thus x comes out to be 20 ml.

c) As at equivalence point all the HCl will be neutralized by the base to form water and salt hence no acid and base will be left only salt and water will be present in the solution thus the pH will be neutal. As the temperature is not specified we can not mention the exact pH. But if we were to take the room temperature then we can say it as 7 as water is neutral at pH 7 at room temperature which is considered as 25 degree celcius. As pH is temperature dependent.