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Hunter-Best [27]
3 years ago
13

In a hospital laboratory, a 10.0 mL sample of gastric juice (predominantly HCl), obtained several hours after a meal, was titrat

ed 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?
Chemistry
1 answer:
LiRa [457]3 years ago
3 0

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

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