Question

If 197 mg of HCL is added to 1 L of solution with a pH of 3, what is the final pH? HCL ↔ H+ + CL-

Answer 1

Given 197mg of HCL = 0.197g of HCL

it is aaded to 1L solution of pH 3

Molecular weight of Hydrogen H = 1g

Molecular weight of Chlorine = 35.46g

• First we will calculate concentration of H+ ions in HCL

HCL ↔ H+ + CL-

36.46g 1g 35.46g

36.46 g of HCL give 1g of H+ ions

1g of HCL will give (1/36.46)g of  H+ ions

Therefore, 0.197 g of HCL will give 0.197*(1/36.46)g = 0.005403g of  H+ ions

Moles of H+ ions = given mass/ molar mass

= 0.005403/1 = 0.005403 moles

Now, the solution has pH = 3

which means it has hydrogen concentration of 10-3 moles/litre

Total Volume of solution = 1 Litre

[ Here we are neglecting the volume contibuted by HCL as it has a density of 1.2g/mL, if we calculate volume of HCL , we will get as (0.197*10-3) / 1.2 = 0.0985 * 10-6 litre, which is ver small compared to 1L]

Total moles of H+ ions = 10-3 + 0.005403 = 0.001+0.005403 = 0.006403 moles

H+ concentration = Total moles/ Total volume in litres = 0.006403 / 1 = 0.006403 moles/litre

Now, we know that

pH = -log10[0.006403] = 2.19

Here, we can observe that pH is much lower because the HCL part has negligible contibution in volume of resulting solution