Question

(a) Sketch the titration curve for an alkalinity test in which the water sample had an initial pH of 6.35. In that test, 15 mL of 0.1 N acid was added to a 150-mL sample.

(b) Calculate the alkalinity in meq/L and mg/L as CaCO₃. Answer: 10 meq/L, 500 mg/L as CaCO₃

_{I know the answer is correct but I am having a problem on a few of the steps, please show all work for a "thumbs up"! Thanks!}

Answer 1

I will do part b) here, cause I'm not very handy in sketching titration curve, I apologize for that. However I can help you with part b) I think I know where's your mistake, so let me put this step by step.

We know that the concentration of the acid is 0.1 N with a volume of 15 mL, added to a volume of base of 150 mL. Assuming they both have a 1:1 relation, we can use the following equation, to determine the volume to reach the equivalence point:

N_aV_a = N_bV_b

So the first step is solve for N_b here. This will give us the alkalinity in units of N (eq/L):

N_b = N_aV_a / V_b = N_a (V_a/V_b)

N_b = 0.150 N (15 mL/150 mL)

N_b = 0.010 N

Now, to get the meq/L you just need to multiply this value by 1000:

N_b = 0.010 eq/L * 1000 meq/eq = 10 meq/L

As a third step, to convert this to units of ppm (mg/L) we need the value of the equivalence weight of CaCO₃ and that's all. First, let's get the molecular weight of CaCO₃:

MW = 40.08 + 12.01 + 3x16 = 100.09 g/mol

To get the Equivalence weight we need to divide this MW by the number of electrons involved in the compound, cause CaCO₃ is a ternary salt, If it were a base, it will be divided by the number of OH ions. But in this case, the carbonate ion has a charge of 2- so:

EW = 100.09 / 2 = 50.045 g/eq

Finally, the alkalinity in mg/L

N_b = 0.010 eq/L * 50.045 g/eq * 1000 mg/g = 500.45 mg/L

Hope this helps, and once again I apologize for not provide part a).