Question

65.24 ml of 0.12 m hcl reacts with 40.00 ml of nh₄oh solution to reach the endpoint using bromothymol blue as an indicator. what is the molar concentration of the ammonium hydroxide? nh₄oh + hcl → nh₄cl + h₂o ? m nh₄oh

Explanation:

Step1: Recall the formula for titration

In acid - base titration, at the endpoint, the moles of acid are equal to the moles of base. The formula is \(M_1V_1 = M_2V_2\) where \(M_1\) and \(V_1\) are the molarity and volume of the acid (HCl) and \(M_2\) and \(V_2\) are the molarity and volume of the base (\(NH_4OH\)). Here, \(M_1 = 0.12\space M\), \(V_1=65.24\space mL\), \(V_2 = 40.00\space mL\) and we need to find \(M_2\).

Step2: Rearrange the formula to solve for \(M_2\)

From \(M_1V_1=M_2V_2\), we can get \(M_2=\frac{M_1V_1}{V_2}\)

Step3: Substitute the values into the formula

Substitute \(M_1 = 0.12\space M\), \(V_1 = 65.24\space mL\) and \(V_2=40.00\space mL\) into the formula:
\(M_2=\frac{0.12\space M\times65.24\space mL}{40.00\space mL}\)
First, calculate the numerator: \(0.12\times65.24 = 7.8288\)
Then, divide by the denominator: \(\frac{7.8288}{40.00}=0.19572\space M\)

Answer:

\(0.19572\) (or approximately \(0.20\) if rounded to two significant figures, but based on the given values, the more accurate value is \(0.19572\))