a solution of hydrochloric acid of unknown concentration was titrated with .21 M NaOH. if a 75 ml sample of the HCl solution required exactly 13ml of the NaOH solution to reach the equivalence point, what was the ph of the HCl solution

Respuesta :

Answer:

pH of HCl solution is 1.44

Explanation:

NaOH is a monoprotic base and HCl is a monobasic acid.

Neutralization reaction: [tex]NaOH+HCl\rightarrow NaCl+H_{2}O[/tex]

According to balanced reaction, 1 mol of NaOH neutralizes 1 mol of HCl.

Number of moles of NaOH in 13 mL of 0.21 M NaOH

= [tex]\frac{0.21}{1000}\times 13mol=0.00273mol[/tex]

let's assume concentration of HCl is C (M)

Then, number of moles of HCl in 75 mL of C (M) HCl solution

= [tex]\frac{75}{1000}\times Cmol=\frac{75C}{1000}mol[/tex]

So, we can write, [tex]\frac{75C}{1000}=0.00273[/tex]

or, [tex]C=0.0364[/tex]

1 mol of HCl contains 1 mol of [tex]H^{+}[/tex]

So, concentration of [tex]H^{+}[/tex] in 0.0364 M HCl, [tex][H^{+}]=0.0364M[/tex]

Hence, [tex]pH=-log[H^{+}]=-log(0.0364)=1.44[/tex]