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?

The pH of the gastric juice is 1.14

We'll begin by calculating the molarity of the HCl needed for the reaction.

HCl + NaOH —> NaCl + H₂O

From the balanced equation above,

The mole ratio of the acid, HCl (nA) = 1

The mole ratio of the base, NaOH (nB) = 1

Molarity of base, NaOH (Mb) = 0.1 M

Volume of base, NaOH (Vb) = 7.2 mL

Volume of acid, HCl (Va) = 10 mL

MaVa / MbVb = nA/nB

(Ma × 10) / (0.1 × 7.2) = 1

(Ma × 10) / 0.72 = 1

Cross multiply

Ma × 10 = 0.72

Divide both side by 10

Ma = 0.72 / 10

HCl (aq) —> H⁺(aq) + Cl¯(aq)

From the balanced equation above,

1 mole of HCl contains 1 mole of H⁺.

Therefore,

0.072 M HCl will also contain 0.072 M H⁺

Hydrogen ion concentration, [H⁺] = 0.072 M

pH = –Log [H⁺]

pH = –Log 0.072

Thus, the pH of the gastric juice is 1.14

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1.14

Explanation:

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

where,

are the n-factor, molarity and volume of acid which is

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

We are given:

Putting values in above equation, we get:

To calculate pH of gastric juice:

molarity of = 0.072

Thus the pH of the gastric juice is 1.14

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