Question

In a hospital laboratory, a 10.0 mL sample of gastric juice (predominantly HCl), obtained several hours after a meal, was titrated with 0.1 M NaOH to neutrality; 7.2 mL of NaOH was required. The patient’s stomach contained no ingested food or drink, thus assume that no buffers were present. What was the pH of the gastric juice?

Answer 1

Answer: 1.14

Explanation:

$HCl+NaOH\rightarrow NaCl+H_2O$

To calculate the molarity 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 $HCl$

are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?\\V_1=10.0mL\\n_2=1\\M_2=0.1M\\V_2=7.2mL$

Putting values in above equation, we get:

$1\times M_1\times 10.0=1\times 0.1\times 7.2\\\\M_1=0.072M$

To calculate pH of gastric juice:

molarity of $H^+$ = 0.072

$pH=-log[H^+]$

$pH=-log(0.072)=1.14$

Thus the pH of the gastric juice is 1.14