Question

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

Answer 1

Answer:

pH of HCl solution is 1.44

Explanation:

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

Neutralization reaction: $NaOH+HCl\rightarrow NaCl+H_{2}O$

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

= $\frac{0.21}{1000}\times 13mol=0.00273mol$

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

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

= $\frac{75}{1000}\times Cmol=\frac{75C}{1000}mol$

So, we can write, $\frac{75C}{1000}=0.00273$

or, $C=0.0364$

1 mol of HCl contains 1 mol of $H^{+}$

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

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