Question

What is the hydronium-ion concentration of a solution formed by combining 400. mL of 0.21 M HNO3 with 600. mL of 0.11 M NaOH at 25 degrees celcius?

A) 0.082 M

B) 6.7 *10^-13

C) 0.21 M

D) 0.018 M

E) 0.11 M

Answer 1

V1 = 400 ml

M1 = 0.21 HNO3

V2 = 600 ml

M = 0.1 M NaOH

HNO3 + NaOH ---> H2O + NaNO3(aq)

Both are strong acid/bases

Calculate mols of each

M1*V1 = 0.4*.21 = 0.084 mol of HNO3

M2*V2 = 0.6*.11 = 0.066 mol of NaOH

There is more aci than base, expect acidic solution

0.066 mol of HNO3 react with 0.066 mol of NaOH to form 0.066 mol of Water and salt

0.084-0.066 = 0.018 mol of HNO3

Recalculate concentration since volume has changed

Vt = v1+V2 = 400 ml + 600 ml = 1000 ml or 1 L

M = mol/L = 0.018 / 1 L = 0.018 M

1 mol of H3O+ present per each mol of HNO3

[H3O+] = 0.018... choose d)