Answer: 1.14
Explanation:
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
To calculate the molarity of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HCl[/tex]
are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=1\\M_1=?\\V_1=10.0mL\\n_2=1\\M_2=0.1M\\V_2=7.2mL[/tex]
Putting values in above equation, we get:
[tex]1\times M_1\times 10.0=1\times 0.1\times 7.2\\\\M_1=0.072M[/tex]
To calculate pH of gastric juice:
molarity of [tex]H^+[/tex] = 0.072
[tex]pH=-log[H^+][/tex]
[tex]pH=-log(0.072)=1.14[/tex]
Thus the pH of the gastric juice is 1.14
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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