Question

Phenolphthalein indicator was added, and the solution in the flask was titrated with 0.215M NaOH until the indicator just turned pink(pH=8-9). The exact value of NaOH required was 22.75mL. Calculate the concentration of HCl in the original 10.00 mL sample. CAN SOMEONE PLEASE EXPLAIN HOW TO DO IT??

Answer 1

In titration, the moles of acid equal moles of base. You were given that 22.75ml of 0.215M NaOH is used, so calculate the number of moles of that base the experiment used in total. After that because you know mol base = mol acid, whatever amount of base you use must be the total amount of acid present in the solution. You were given the volume of the acid, and you have just found the total mols of acid. Using these two information, solve for the concentration. And one more thing, even though I'm pretty sure it won't affect your answer, you should always convert things to the proper units. Since the concentration we're talking about in this problem is molarity, which has the unit mol/L, you should always have all of your numbers in these units. It just make it simpler and will not confuse you

Answer 2

Answer:

0489 M is the concentration of HCl in the original 10.00 mL sample.

Explanation:

$HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)$

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1, M_1=?,V_1=10.00 mL$

$n_2=1,M_2=0.215 M,V_2-=22.75 mL$

$n_1M_1V_1=n_2M_2V_2$

$M_1=\frac{n_2M_2V_2}{n_1\times V_1}$

$=\frac{1\times 0.215 M\times 22.75 mL}{1\times 10.00 mL}=0.489 M$

0489 M is the concentration of HCl in the original 10.00 mL sample.