# Can u please check. Would this answer be right?

#1) In reference to the information given above, determine the following.
A 0.10 mol/L solution of H2Te(aq) is prepared and found to have a pH= 1.85. Calculate the Ka value for H2Te(aq).

The Ka value for H2Te is 0.002279mol2/L2 .

#1) In reference to the information given above, determine the following.
A 0.10 mol/L solution of H2Te(aq) is prepared and found to have a pH= 1.85. Calculate the Ka value for H2Te(aq).

The Ka value for H2Te is 0.002279mol2/L2 .

What is the information above? Is it pertinent to the problem. And do you want Ka (which I consider as k1k2) or do you want k1?

9 months ago

## The information given above states that a 0.10 mol/L solution of H2Te(aq) has a pH of 1.85.

To calculate the Ka value for H2Te(aq), we need to use the equation that relates pH to pKa:

pH = pKa + log([A-]/[HA])

Where [A-] is the concentration of the conjugate base of the acid and [HA] is the concentration of the acid.

Now, we can rearrange the equation to solve for pKa:

pKa = pH - log([A-]/[HA])

In this case, H2Te is the acid and Te- is its conjugate base. Since the solution didn't provide the concentration of [A-]/[HA], we cannot directly calculate pKa.

However, if we assume that H2Te is a weak acid (which is typically the case for these types of problems), we can make an approximation that [A-] is negligibly small compared to [HA]. This assumption allows us to ignore the concentration of the conjugate base in the equation, simplifying it to:

pKa = pH

Now, we can substitute the pH value of 1.85 into the equation to get the pKa value.

However, it's important to note that pKa is the negative logarithm of the Ka value, so to get the Ka value, we need to take the antilog of pKa.

To summarize, based on the information provided, the pKa value can be approximated as 1.85, and the Ka value for H2Te(aq) can be calculated by taking the antilog of 1.85.