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a patient is suspected of having low stomach acid, a condition known as hypochloridia. to determine whether the patient has this condition, her doctors take a 20.00 ml sample of her gastric juices and titrate the sample with 1.72×10−4 m koh. the gastric juice sample required 17.2 ml of the koh titrant to neutralize it. calculate the ph of the gastric juice sample. assume the sample contained no ingested food or drink which might otherwise interfere with the titration

Answer :

The answer would be 5.1

Volume of KOH, V1 = 17.2mL

Volume of HCI, V2 = 20mL

Molarity of KOH, M1 = 1.72 * 10 ^ - 4 * M

The given titration is between KOH and gastric juice (HCI).

The molarity of HCI can be calculated using the relation,

M1 * V1 =M2 * V2 ---------- (1)

M2 = M1 * V1 / V2

Where,

M1 is the molarity of KOH.

M2 is the molarity of HCI.

V1 is the volume of KOH.

V2 is the volume of HCI.

Substituting the values in equation (1),

M2 = 1.72 * 10^-4 * 17.2 / 20.00

M2 = 1.49 * 10^ -5 g

The pH of gastric juice (HCI) can be calculated using the

equation, pH=-log[H^ + ] -------- (II)

Since, HCI is a strong acid, it dissociates as follows, HCI H+CI

Therefore,

[HCl]=[H^ + ]=[Cl^ - ]= 1.49 * 10 ^ - 5 * M

Substituting this in equation (II),

pH =-log[H^ + ] = - log[1.49 * 10 ^ - 5] = - log(1.49) + 5 * log(10)

The value of pH is calculated as,

pH = - log(4.78) + 5 * log(10) = -0.17314 +5 = 5.1

Hence, pH of gastric juice is 5.1

To know more about pH here :

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